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== 4.4#h(0.6em)Properties of Logarithms


    
      
=== Introduction


      Because logarithms are actually exponents, they have several properties that can be derived from the laws of exponents. Here are the laws we will need at present.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Laws of Exponents]


        + To multiply two powers with the same base, add the exponents and leave the base unchanged.
			#math.equation(block: true, alt: "a to the power m times a to the power n equals a to the power m plus n")[$a^(m) ⋅ a^(n) = a^(m + n)$]
+ To divide two powers with the same base, subtract the exponents and leave the base unchanged.
			#math.equation(block: true, alt: "the fraction a to the power m over a to the power n equals a to the power m minus n")[$frac(a^(m), a^(n)) = a^(m − n)$]
+ To raise a power to a power, keep the same base and multiply the exponents.
			#math.equation(block: true, alt: "open parenthesis a to the power m close parenthesis to the power n equals a to the power m n")[$attach(( a^(m) ), t: n) = a^(m n)$]
]


      Each of these laws corresponds to one of three properties of logarithms.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Properties of Logarithms]


        If #math.equation(block: false, alt: "x")[$x$], #math.equation(block: false, alt: "y")[$y$], #math.equation(block: false, alt: "b greater than 0")[$b > 0$], and #math.equation(block: false, alt: "b not equal to 1")[$b ≠ 1$], then


        + #math.equation(block: false, alt: "log sub b ⁡ open parenthesis x y close parenthesis equals log sub b ⁡ open parenthesis x close parenthesis plus log sub b ⁡ open parenthesis y close parenthesis")[$log_(b) ( x y ) = log_(b) ( x ) + log_(b) ( y )$]
+ #math.equation(block: false, alt: "log sub open parenthesis b close parenthesis ⁡ the fraction x over y equals log sub b ⁡ open parenthesis x close parenthesis minus log sub b ⁡ open parenthesis y close parenthesis")[$log_(( b )) display(frac(x, y)) = log_(b) ( x ) − log_(b) ( y )$]
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis x to the power k close parenthesis equals k log sub b ⁡ open parenthesis x close parenthesis")[$log_(b) ( x^(k) ) = k log_(b) ( x )$]
]


      We will consider proofs of the three properties of logarithms in the Homework problems. For now, study the examples below, keeping in mind that a logarithm is an exponent.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
+ Property (1):
        #math.equation(block: true, alt: "log sub 2 ⁡ open parenthesis 32 close parenthesis equals log sub 2 ⁡ open parenthesis 4 times 8 close parenthesis, equals log sub 2 ⁡ 4 plus log sub 2 ⁡ 8 because 2 to the power 5 equals 2 squared times 2 cubed; 5, equals 2 plus 332 equals 4 times 8")[$log_(2) ( 32 ) = bold(log)_(2) ( 4 ⋅ 8 ) & = bold(log)_(2) 4 + bold(log)_(2) 8 " because " 2^(5) = 2^(2) ⋅ 2^(3) \ 5 & = 2 + 3 " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " 32 = 4 ⋅ 8$]
+ Property (2):
        #math.equation(block: true, alt: "log sub 2 ⁡ open parenthesis 8 close parenthesis equals log sub 2 ⁡ open parenthesis the fraction 16 over 2 close parenthesis, equals log sub 2 ⁡ open parenthesis 16 close parenthesis minus log sub 2 ⁡ open parenthesis 2 close parenthesis because 2 cubed equals the fraction 2 to the power 4 over 2 to the power 1; 3, equals 4 minus 18 equals the fraction 16 over 2")[$log_(2) ( 8 ) = bold(log)_(2) ( frac(16, 2) ) & = bold(log)_(2) ( 16 ) − bold(log)_(2) ( 2 ) " because " 2^(3) = frac(2^(4), 2^(1)) \ 3 & = 4 − 1 " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " 8 = frac(16, 2)$]
+ Property (3):
         Property (3):
        #math.equation(block: true, alt: "log sub 2 ⁡ 64 equals log sub 2 ⁡ open parenthesis 4 close parenthesis cubed, equals 3 log sub 2 ⁡ open parenthesis 4 close parenthesis because open parenthesis 2 squared close parenthesis cubed equals 2 to the power 6; 6, equals 3 times 2 open parenthesis 4 close parenthesis cubed equals 64")[$log_(2) 64 = bold(log)_(2) ( 4 )^(3) & = 3 bold(log)_(2) ( 4 ) " because " attach(( 2^(2) ), t: 3) = 2^(6) \ 6 & = 3 ⋅ 2 " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " ( 4 )^(3) = 64$]
]


      
        Which statement is true?


        \_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "log ⁡ open parenthesis 8 y close parenthesis equals log ⁡ open parenthesis 8 close parenthesis plus log ⁡ open parenthesis y close parenthesis")[$log ( 8 y ) = log ( 8 ) + log ( y )$]
]


      
      
        Which statement is true?


        + #math.equation(block: false, alt: "log ⁡ open parenthesis 8 plus y close parenthesis equals log ⁡ open parenthesis 8 close parenthesis plus log ⁡ open parenthesis y close parenthesis")[$log ( 8 + y ) = log ( 8 ) + log ( y )$]
+ #math.equation(block: false, alt: "log ⁡ open parenthesis 8 y close parenthesis equals log ⁡ open parenthesis 8 close parenthesis plus log ⁡ open parenthesis y close parenthesis")[$log ( 8 y ) = log ( 8 ) + log ( y )$]
+ #math.equation(block: false, alt: "log ⁡ open parenthesis 8 y close parenthesis equals log ⁡ open parenthesis 8 close parenthesis times log ⁡ open parenthesis y close parenthesis")[$log ( 8 y ) = log ( 8 ) ⋅ log ( y )$]
+ #math.equation(block: false, alt: "log ⁡ open parenthesis the fraction 8 over y close parenthesis equals the fraction log ⁡ open parenthesis 8 close parenthesis over log ⁡ open parenthesis y close parenthesis")[$log ( display(frac(8, y)) ) = display(frac(log ( 8 ), log ( y )))$]


      
    
    
      
=== Using the Properties of Logarithms


      Of course, these properties are useful not so much for computing logs but rather for simplifying expressions that contain variables. We will use them to solve exponential equations. But first, we will practice applying the properties. In the following Example, we rewrite one log in terms of simpler logs.


      #examplebox("Example 1")[][
Simplify #math.equation(block: false, alt: "log sub b ⁡ open parenthesis the square root of x y close parenthesis")[$" " " " log_(b) ( sqrt(x y) )$].


        
        #solutionbox[
First, we write #math.equation(block: false, alt: "the square root of x y")[$sqrt(x y)$] using a fractional exponent:


          #math.equation(block: true, alt: "log sub b ⁡ open parenthesis x y close parenthesis equals log sub b ⁡ open parenthesis x y close parenthesis to the power 1 / 2")[$log_(b) ( x y ) = log_(b) ( x y )^(1 / 2)$]
          Then we apply Property (3) to rewrite the exponent as a coefficient:


          #math.equation(block: true, alt: "log sub b ⁡ open parenthesis x y close parenthesis to the power 1 / 2 equals the fraction 1 over 2 log sub b ⁡ open parenthesis x y close parenthesis")[$log_(b) ( x y )^(1 / 2) = frac(1, 2) log_(b) ( x y )$]
          Finally, by Property (1) we write the log of a product as a sum of logs:


          #math.equation(block: true, alt: "the fraction 1 over 2 open parenthesis log sub b ⁡ open parenthesis x y close parenthesis close parenthesis equals the fraction 1 over 2 open parenthesis log sub b ⁡ open parenthesis x close parenthesis plus log sub b ⁡ open parenthesis y close parenthesis close parenthesis")[$frac(1, 2) ( log_(b) ( x y ) ) = frac(1, 2) ( log_(b) ( x ) + log_(b) ( y ) )$]
          Thus, #math.equation(block: false, alt: "log sub b ⁡ open parenthesis the square root of x y close parenthesis equals the fraction 1 over 2 open parenthesis log sub b ⁡ open parenthesis x close parenthesis plus log sub b ⁡ open parenthesis y close parenthesis close parenthesis")[$" " " " log_(b) ( sqrt(x y) ) = display(frac(1, 2)) ( log_(b) ( x ) + log_(b) ( y ) )$].
]
]


      
        Simplify #math.equation(block: false, alt: "log sub b ⁡ open parenthesis the fraction x over y squared close parenthesis")[$log_(b) ( display(frac(x, y^(2))) )$].


        Answer: #math.equation(block: false, alt: "log sub b")[$log_(b)$]\_\_\_\_\_#math.equation(block: false, alt: "plus")[$+$]\_\_\_\_\_#math.equation(block: false, alt: "times log sub b")[$⋅ log_(b)$]\_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "log sub b ⁡ x minus 2 log sub b ⁡ y")[$log_(b) x − 2 log_(b) y$]
]


      
      
        Simplify #math.equation(block: false, alt: "log sub b ⁡ open parenthesis the fraction x over y squared close parenthesis")[$log_(b) ( display(frac(x, y^(2))) )$].


        #solutionbox[
#math.equation(block: true, alt: "log sub b ⁡ open parenthesis x close parenthesis minus 2 log sub b ⁡ open parenthesis y close parenthesis")[$log_(b) ( x ) − 2 log_(b) ( y )$]
]


      
      #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Be careful when using the properties of logarithms! Compare the statements below:


        + #math.equation(block: false, alt: "log sub b ⁡ open parenthesis 2 x close parenthesis equals log sub b ⁡ open parenthesis 2 close parenthesis plus log sub b ⁡ open parenthesis x close parenthesis by Property 1,")[$log_(b) ( 2 x ) = log_(b) ( 2 ) + log_(b) ( x ) " " " " " " " " " by Property 1,"$] #linebreak() but #linebreak() #math.equation(block: false, alt: "log sub b ⁡ open parenthesis 2 plus x close parenthesis not equal to log sub b ⁡ open parenthesis 2 close parenthesis plus log sub b ⁡ open parenthesis x close parenthesis")[$log_(b) ( 2 + x ) ≠ log_(b) ( 2 ) + log_(b) ( x )$] #linebreak()
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis the fraction x over 5 close parenthesis equals log sub b ⁡ open parenthesis x close parenthesis minus log sub b ⁡ open parenthesis 5 close parenthesis by Property 2,")[$log_(b) ( display(frac(x, 5)) ) = log_(b) ( x ) − log_(b) ( 5 ) " " " " " " " " " by Property 2,"$] #linebreak() but #linebreak() #math.equation(block: false, alt: "log sub b ⁡ open parenthesis the fraction x over 5 close parenthesis not equal to the fraction log sub b ⁡ open parenthesis x close parenthesis over log sub b ⁡ open parenthesis 5 close parenthesis")[$log_(b) ( display(frac(x, 5)) ) ≠ display(frac(log_(b) ( x ), log_(b) ( 5 )))$]
]


      We can also use the properties of logarithms to combine sums and differences of logarithms into one logarithm.


      #examplebox("Example 2")[][
Express #math.equation(block: false, alt: "3 open parenthesis log sub b ⁡ open parenthesis x close parenthesis minus log sub b ⁡ open parenthesis y close parenthesis close parenthesis")[$" " " " 3 ( log_(b) ( x ) − log_(b) ( y ) )$] as a single logarithm with a coefficient of #math.equation(block: false, alt: "1")[$1$].


        
        #solutionbox[
We begin by applying Property (2) to combine the logs.


          #math.equation(block: true, alt: "3 open parenthesis log sub b ⁡ open parenthesis x close parenthesis minus log sub b ⁡ open parenthesis y close parenthesis close parenthesis equals 3 log sub b ⁡ open parenthesis the fraction x over y close parenthesis")[$3 ( log_(b) ( x ) − log_(b) ( y ) ) = 3 log_(b) ( frac(x, y) )$]
          Then, using Property (3), we replace the coefficient #math.equation(block: false, alt: "3")[$3$] by an exponent #math.equation(block: false, alt: "3")[$3$].


          #math.equation(block: true, alt: "3 log sub b ⁡ open parenthesis the fraction x over y close parenthesis equals log sub b ⁡ open parenthesis open parenthesis the fraction x over y close parenthesis cubed close parenthesis")[$3 log_(b) ( frac(x, y) ) = log_(b) ( attach(( frac(x, y) ), t: 3) )$]
]
]


      
        Express #math.equation(block: false, alt: "2 log sub b ⁡ x plus 4 log sub b ⁡ open parenthesis x plus 3 close parenthesis")[$2 log_(b) x + 4 log_(b) ( x + 3 )$] as a single logarithm with a coefficient of #math.equation(block: false, alt: "1")[$1$].


        Answer: #math.equation(block: false, alt: "log sub b")[$log_(b)$]\_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "log sub b ⁡ x squared open parenthesis x plus 3 close parenthesis to the power 4")[$log_(b) x^(2) ( x + 3 )^(4)$]
]


      
      
        Express #math.equation(block: false, alt: "2 log sub b ⁡ open parenthesis x close parenthesis plus 4 log sub b ⁡ open parenthesis x plus 3 close parenthesis")[$" " 2 log_(b) ( x ) + 4 log_(b) ( x + 3 ) " "$] as a single logarithm with a coefficient of #math.equation(block: false, alt: "1")[$1$].


        #solutionbox[
#math.equation(block: true, alt: "log sub b ⁡ open parenthesis x squared open parenthesis x plus 3 close parenthesis to the power 4 close parenthesis")[$log_(b) ( x^(2) ( x + 3 )^(4) )$]
]


      
      
        Which expression is equivalent to #math.equation(block: false, alt: "3 log sub 2 ⁡ open parenthesis H close parenthesis minus log sub 2 ⁡ open parenthesis T close parenthesis")[$3 log_(2) ( H ) − log_(2) ( T )$]?


        \_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "log sub 2 ⁡ open parenthesis the fraction H cubed over T close parenthesis")[$log_(2) ( display(frac(H^(3), T)) )$]
]


      
      
        Which expression is equivalent to #math.equation(block: false, alt: "3 log sub 2 ⁡ open parenthesis H close parenthesis minus log sub 2 ⁡ open parenthesis T close parenthesis")[$3 log_(2) ( H ) − log_(2) ( T )$]?


        + #math.equation(block: false, alt: "3 log sub 2 ⁡ open parenthesis H minus T close parenthesis")[$3 log_(2) ( H − T )$]
+ #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis H cubed minus T close parenthesis")[$log_(2) ( H^(3) − T )$]
+ #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis the fraction H cubed over T close parenthesis")[$log_(2) ( display(frac(H^(3), T)) )$]
+ #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis open parenthesis the fraction H over T close parenthesis cubed close parenthesis")[$log_(2) ( attach(( display(frac(H, T)) ), t: 3) )$]


      
    
    
      
=== Solving Exponential Equations


      By using Property (3), we can now solve exponential equations in which the base is not #math.equation(block: false, alt: "10")[$10$]. For example, to solve the equation


      #math.equation(block: true, alt: "5 to the power x equals 7")[$5^(x) = 7$]
      we could rewrite the equation in logarithmic form to obtain the exact solution


      #math.equation(block: true, alt: "x equals log sub 5 ⁡ open parenthesis 7 close parenthesis")[$x = log_(5) ( 7 )$]
      However, we cannot evaluate #math.equation(block: false, alt: "log sub 5 ⁡ open parenthesis 7 close parenthesis")[$log_(5) ( 7 )$]; there is no log base #math.equation(block: false, alt: "5")[$5$] button on the calculator. If we want a decimal approximation for the solution, we begin by taking the base #math.equation(block: false, alt: "10")[$10$] logarithm of both sides, even though the base of the power is not #math.equation(block: false, alt: "10")[$10$]. This gives us


      #math.equation(block: true, alt: "log sub 10 ⁡ open parenthesis 5 to the power x close parenthesis equals log sub 10 ⁡ open parenthesis 7 close parenthesis")[$log_(10) ( 5^(x) ) = log_(10) ( 7 )$]
      Then we use Property (3) to rewrite the left side as


      #math.equation(block: true, alt: "x log sub 10 ⁡ open parenthesis 5 close parenthesis equals log sub 10 ⁡ open parenthesis 7 close parenthesis")[$x log_(10) ( 5 ) = log_(10) ( 7 )$]
      Note how using Property (3) allows us to solve the equation: The variable, #math.equation(block: false, alt: "x")[$x$], is no longer in the exponent, and it is multiplied by a constant, #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 5 close parenthesis")[$log_(10) ( 5 )$]. To finish the solution, we divide both sides by #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 5 close parenthesis")[$log_(10) ( 5 )$] to get


      #math.equation(block: true, alt: "x equals the fraction log sub 10 ⁡ open parenthesis 7 close parenthesis over log sub 10 ⁡ open parenthesis 5 close parenthesis")[$x = frac(log_(10) ( 7 ), log_(10) ( 5 ))$]
      On your calculator, enter the sequence


      LOG #math.equation(block: false, alt: "7")[$7$] ) ÷ LOG #math.equation(block: false, alt: "5")[$5$] ) ENTER


      to find that #math.equation(block: false, alt: "x approximately equals 1.2091")[$x ≈ 1.2091$].


      #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Do not confuse the expression #math.equation(block: false, alt: "the fraction log sub 10 ⁡ open parenthesis 7 close parenthesis over log sub 10 ⁡ open parenthesis 5 close parenthesis")[$" " display(frac(log_(10) ( 7 ), log_(10) ( 5 ))) " "$] with #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis the fraction 7 over 5 close parenthesis")[$" " log_(10) ( display(frac(7, 5)) )$]; they are not the same!


        Property (2) allows us to simplify #math.equation(block: false, alt: "log ⁡ open parenthesis the fraction x over y close parenthesis")[$" " log ( display(frac(x, y)) )$], but not #math.equation(block: false, alt: "the fraction log ⁡ open parenthesis x close parenthesis over log ⁡ open parenthesis y close parenthesis")[$" " display(frac(log ( x ), log ( y )))$]. We cannot rewrite #math.equation(block: false, alt: "the fraction log sub 10 ⁡ open parenthesis 7 close parenthesis over log sub 10 ⁡ open parenthesis 5 close parenthesis")[$" " display(frac(log_(10) ( 7 ), log_(10) ( 5 )))$], so we must evaluate it as #math.equation(block: false, alt: "open parenthesis log ⁡ 7 close parenthesis / open parenthesis log ⁡ open parenthesis 5 close parenthesis close parenthesis")[$" " ( log 7 ) / ( log ( 5 ) )$]. You can check on your calculator that


        #math.equation(block: true, alt: "the fraction log sub 10 ⁡ open parenthesis 7 close parenthesis over log sub 10 ⁡ open parenthesis 5 close parenthesis not equal to log sub 10 ⁡ open parenthesis the fraction 7 over 5 close parenthesis equals log sub 10 ⁡ open parenthesis 1.4 close parenthesis")[$display(frac(log_(10) ( 7 ), log_(10) ( 5 ))) ≠ log_(10) ( display(frac(7, 5)) ) = log_(10) ( 1.4 )$]
        .
]


      
        How can we solve #math.equation(block: false, alt: "4 to the power x equals 60")[$4^(x) = 60$]?


        \_\_\_\_\_


        #solutionbox[
Take log base 10 of both sides.
]


      
      
        How can we solve #math.equation(block: false, alt: "4 to the power x equals 60")[$" " 4^(x) = 60$]?


        + Divide both sides by 4.
+ Take log base 10 of both sides.
+ Take the fourth root of 60.
+ This equation has no solution.


      
      #examplebox("Example 3")[][
Solve #math.equation(block: false, alt: "1640 equals 80 times 6 to the power 0.03 x")[$" " " " 1640 = 80 ⋅ 6^(0.03 x)$]


        
        #solutionbox[
First we divide both sides by #math.equation(block: false, alt: "80")[$80$] to obtain


          #math.equation(block: true, alt: "20.5 equals 6 to the power 0.03 x")[$20.5 = 6^(0.03 x)$]
          Next, we take the base #math.equation(block: false, alt: "10")[$10$] logarithm of both sides of the equation and use Property (3) of logarithms to get


          #math.equation(block: true, alt: "log sub 10 ⁡ open parenthesis 20.5 close parenthesis equals log sub 10 ⁡ open parenthesis 6 to the power 0.03 x close parenthesis equals 0.03 x log sub 10 ⁡ open parenthesis 6 close parenthesis")[$log_(10) ( 20.5 ) = log_(10) ( 6^(0.03 x) ) = 0.03 x log_(10) ( 6 )$]
          On the right side of the equation, #math.equation(block: false, alt: "x")[$x$] is multiplied by two constants, #math.equation(block: false, alt: "0.03")[$0.03$] and #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 6 close parenthesis")[$log_(10) ( 6 )$]. So, to solve for #math.equation(block: false, alt: "x")[$x$] we must divide both sides of the equation by #math.equation(block: false, alt: "0.03 log sub 10 ⁡ open parenthesis 6 close parenthesis")[$0.03 log_(10) ( 6 )$]. We use a calculator to evaluate the answer:


          #math.equation(block: true, alt: "x equals the fraction log sub 10 ⁡ open parenthesis 20.5 close parenthesis over 0.03 log sub 10 ⁡ open parenthesis 6 close parenthesis approximately equals 56.19")[$x = frac(log_(10) ( 20.5 ), 0.03 log_(10) ( 6 )) ≈ 56.19$]
          (On your calculator, remember to enclose the denominator, #math.equation(block: false, alt: "0.03 log sub 10 ⁡ open parenthesis 6 close parenthesis")[$0.03 log_(10) ( 6 )$], in parentheses.)
]
]


      #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
In Example, do not try to simplify


        #math.equation(block: true, alt: "80 times 6 to the power 0.03 x → 480 to the power 0.03 x Incorrect!")[$80 ⋅ 6^(0.03 x) → 480^(0.03 x) " " " " " Incorrect!"$]
        Remember that the order of operations tells us to compute the power #math.equation(block: false, alt: "6 to the power 0.03 x")[$6^(0.03 x)$] before multiplying by #math.equation(block: false, alt: "80")[$80$].
]


      We summarize our method for solving exponential equations as follows.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Steps for Solving Exponential Equations]


        + Isolate the power on one side of the equation.
+ Take the log base #math.equation(block: false, alt: "10")[$10$] of both sides.
+ Simplify by applying Log Property (3).
+ Solve for the variable.
]


      
        Solve #math.equation(block: false, alt: "5 open parenthesis 1.2 close parenthesis to the power 2.5 x equals 77")[$5 ( 1.2 )^(2.5 x) = 77$]


        #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_


        
          #math.equation(block: true, alt: "Divide both sides by 5.")[$"Divide both sides by 5."$]


          #math.equation(block: true, alt: "Take the log of both sides.")[$"Take the log of both sides."$]


          #math.equation(block: true, alt: "Apply Property (3) to simplify the left side.")[$"Apply Property (3) to simplify the left side."$]


          #math.equation(block: true, alt: "Solve for x .")[$"Solve for" " " bold(italic(x)) .$]


        
        #solutionbox[
#math.equation(block: true, alt: "x equals the fraction log ⁡ open parenthesis 15.4 close parenthesis over 2.5 log ⁡ open parenthesis 1.2 close parenthesis approximately equals 5.999")[$x = display(frac(log ( 15.4 ), 2.5 log ( 1.2 ))) ≈ 5.999$]
]


      
      
        Solve #math.equation(block: false, alt: "5 open parenthesis 1.2 close parenthesis to the power 2.5 x equals 77")[$5 ( 1.2 )^(2.5 x) = 77$]


        #solutionbox[
Divide both sides by 5, then take the log of both sides. Simplify the left side and solve for #math.equation(block: false, alt: "x")[$x$] to find
        #math.equation(block: false, alt: "x equals the fraction log ⁡ open parenthesis 15.4 close parenthesis over 2.5 log ⁡ open parenthesis 1.2 close parenthesis approximately equals 5.999")[$" " x = display(frac(log ( 15.4 ), 2.5 log ( 1.2 ))) ≈ 5.999$]
]


      
      
        Explain why the distributive law does not apply to the expression #math.equation(block: false, alt: "log ⁡ open parenthesis a plus b close parenthesis")[$log ( a + b )$].


        \_\_\_\_\_


      
      
        Explain why the distributive law does not apply to the expression #math.equation(block: false, alt: "log ⁡ open parenthesis a plus b close parenthesis")[$log ( a + b )$].


      
    
    
      
=== Applications


      By using the properties of logarithms, we can now solve equations that arise in exponential growth and decay models, no matter what base the exponential function uses.


      #examplebox("Example 4")[][
The population of Silicon City was #math.equation(block: false, alt: "6500")[$6500$] in #math.equation(block: false, alt: "1990")[$1990$] and has been tripling every #math.equation(block: false, alt: "12")[$12$] years. When will the population reach #math.equation(block: false, alt: "75 , 000")[$75 , 000$]?


        
        #solutionbox[
The population of Silicon City grows according to the formula


          #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 6500 times 3 to the power t / 12")[$P ( t ) = 6500 ⋅ 3^(t / 12)$]
          where #math.equation(block: false, alt: "t")[$t$] is the number of years after #math.equation(block: false, alt: "1990")[$1990$]. We want to find the value of #math.equation(block: false, alt: "t")[$t$] for which #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals 75 , 000")[$P ( t ) = 75 , 000$]; that is, we want to solve the equation


          #math.equation(block: true, alt: "6500 times 3 to the power t / 12, equals 75 , 000, Divide both sides by 6500.; 3 to the power t / 12, equals the fraction 150 over 13")[$6500 ⋅ 3^(t / 12) & = 75 , 000 &  & " Divide both sides by 6500." \ 3^(t / 12) & = frac(150, 13)$]
          Now we take the base #math.equation(block: false, alt: "10")[$10$] logarithm of both sides and solve for #math.equation(block: false, alt: "t")[$t$].


          #math.equation(block: true, alt: "log sub 10 ⁡ open parenthesis 3 to the power t / 12 close parenthesis, equals log sub 10 ⁡ open parenthesis the fraction 150 over 13 close parenthesis, Apply Property (3).; the fraction t over 12 log sub 10 ⁡ open parenthesis 3 close parenthesis, equals log sub 10 ⁡ open parenthesis the fraction 150 over 13 close parenthesis, Divide by log sub 10 ⁡ open parenthesis 3 close parenthesis ; multiply by 12.; t, equals the fraction 12 log sub 10 ⁡ open parenthesis the fraction 150 over 13 close parenthesis over log sub 10 ⁡ open parenthesis 3 close parenthesis; approximately equals 26.71")[$log_(10) ( 3^(t / 12) ) & = log_(10) ( frac(150, 13) ) &  & "Apply Property (3)." \ frac(t, 12) log_(10) ( 3 ) & = log_(10) ( frac(150, 13) ) &  & "Divide by " bold(log)_(10) ( 3 ) "; multiply by 12." \ t & = frac(12 log_(10) ( display(frac(150, 13)) ), log_(10) ( 3 )) \  & ≈ 26.71$]
          The population of Silicon City will reach #math.equation(block: false, alt: "75 , 000")[$75 , 000$] about #math.equation(block: false, alt: "27")[$27$] years after #math.equation(block: false, alt: "1990")[$1990$], or in #math.equation(block: false, alt: "2017")[$2017$].
]
]


      
        Traffic on U.S. highways is growing by 2.7% per year. (Source: #emph[Time], Jan. 25, 1999)


        + Write a formula for the volume, #math.equation(block: false, alt: "V")[$V$], of traffic as a function of time, using #math.equation(block: false, alt: "V sub 0")[$V_(0)$] for the current volume.  \[Note: Enter "V0" to get #math.equation(block: false, alt: "V sub 0")[$V_(0)$].\] #linebreak() #math.equation(block: false, alt: "V equals")[$V =$]\_\_\_\_\_
+ How long will it take the volume of traffic to double? #emph[Hint:] Find the value of #math.equation(block: false, alt: "t")[$t$] that gives #math.equation(block: false, alt: "V equals 2 V sub 0")[$V = 2 V_(0)$]. #linebreak() Answer: about \_\_\_\_\_ years


        #solutionbox[
+ #math.equation(block: false, alt: "V open parenthesis t close parenthesis equals V sub 0 open parenthesis 1.027 close parenthesis to the power t")[$V ( t ) = V_(0) ( 1.027 )^(t)$]
+ about #math.equation(block: false, alt: "26")[$26$] years
]


      
      
        Traffic on U.S. highways is growing by 2.7% per year. (Source: #emph[Time], Jan. 25, 1999)


        + Write a formula for the volume, #math.equation(block: false, alt: "V")[$V$], of traffic as a function of time, using #math.equation(block: false, alt: "V sub 0")[$V_(0)$] for the current volume.
+ How long will it take the volume of traffic to double? #emph[Hint:] Find the value of #math.equation(block: false, alt: "t")[$t$] that gives #math.equation(block: false, alt: "V equals 2 V sub 0")[$V = 2 V_(0)$].


        #solutionbox[
+ #math.equation(block: false, alt: "V open parenthesis t close parenthesis equals V sub 0 open parenthesis 1.027 close parenthesis to the power t")[$V ( t ) = V_(0) ( 1.027 )^(t)$]
+ about #math.equation(block: false, alt: "26")[$26$] years
]


      
      
        How does Property (3) of logarithms help us solve exponential equations?


        \_\_\_\_\_


      
      
        How does Property (3) of logarithms help us solve exponential equations?


      
    
    
      
=== Compound Interest


      The amount of money in an account that earns interest compounded annually grows exponentially according to the formula


      #math.equation(block: true, alt: "A open parenthesis t close parenthesis equals P open parenthesis 1 plus r close parenthesis to the power t")[$A ( t ) = P ( 1 + r )^(t)$]
      (See Exponential Growth and Decay to review compound interest.) Many accounts compound interest more frequently than once a year. If the interest is compounded #math.equation(block: false, alt: "n")[$n$] times per year, then in #math.equation(block: false, alt: "t")[$t$] years there will be #math.equation(block: false, alt: "n t")[$n t$] compounding periods, and in each period the account earns interest at a rate of #math.equation(block: false, alt: "the fraction r over n")[$display(frac(r, n))$]. The amount accumulated is given by a generalization of our earlier formula.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Compound Interest]


        The amount #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$] accumulated (principal plus interest) in an account bearing interest compounded #math.equation(block: false, alt: "n")[$n$] times annually is


        #math.equation(block: true, alt: "A open parenthesis t close parenthesis equals P open parenthesis 1 plus the fraction r over n close parenthesis to the power n t")[$A ( t ) = P attach(( 1 + frac(r, n) ), t: n t)$]
        where


        #figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([], [], [#math.equation(block: false, alt: "P")[$P$]], [is the principal invested,]),
  [], [], [#math.equation(block: false, alt: "r")[$r$]], [is the interest rate,],
  [], [], [#math.equation(block: false, alt: "t")[$t$]], [is the time period, in years.],
))
]


      #examplebox("Example 5")[][
Rashad deposited \$#math.equation(block: false, alt: "1000")[$1000$] in an account that pays #math.equation(block: false, alt: "4")[$4$]% interest. Calculate the amount in his account after #math.equation(block: false, alt: "5")[$5$] years if the interest is compounded


          + semiannually
+ quarterly
+ monthly


        
        #solutionbox[
+ #emph[Semiannually] means "twice a year,"" so we use the formula for compound interest with #math.equation(block: false, alt: "P equals 1000")[$P = 1000$], #math.equation(block: false, alt: "r equals 0.04")[$r = 0.04$], #math.equation(block: false, alt: "n equals 2")[$n = 2$], and #math.equation(block: false, alt: "t equals 5")[$t = 5$].
			#math.equation(block: true, alt: "A open parenthesis 5 close parenthesis, equals 1000 open parenthesis 1 plus the fraction 0.04 over 2 close parenthesis to the power 2 open parenthesis 5 close parenthesis; equals 1000 open parenthesis 1.02 close parenthesis to the power 10 equals 1218.99")[$A ( 5 ) & = 1000 attach(( 1 + frac(0.04, 2) ), t: 2 ( 5 )) \  & = 1000 ( 1.02 )^(10) = 1218.99$]
	If interest is compounded semiannually, the balance in the account after #math.equation(block: false, alt: "5")[$5$] years is \$#math.equation(block: false, alt: "1218.99")[$1218.99$].
+ #emph[Quarterly] means "4 times a year,"" so we use the formula for compound interest with #math.equation(block: false, alt: "P equals 1000")[$P = 1000$], #math.equation(block: false, alt: "r equals 0.04")[$r = 0.04$], #math.equation(block: false, alt: "n equals 4")[$n = 4$], and #math.equation(block: false, alt: "t equals 5")[$t = 5$].
			#math.equation(block: true, alt: "A open parenthesis 5 close parenthesis, equals 1000 open parenthesis 1 plus the fraction 0.04 over 4 close parenthesis to the power 4 open parenthesis 5 close parenthesis; equals 1000 open parenthesis 1.01 close parenthesis to the power 20 equals 1220.19")[$A ( 5 ) & = 1000 attach(( 1 + frac(0.04, 4) ), t: 4 ( 5 )) \  & = 1000 ( 1.01 )^(20) = 1220.19$]
	If interest is compounded quarterly, the balance in the account after #math.equation(block: false, alt: "5")[$5$] years is \$#math.equation(block: false, alt: "1220.19")[$1220.19$].
+ There are #math.equation(block: false, alt: "12")[$12$] months in a year, so we use the formula for compound interest with #math.equation(block: false, alt: "P equals 1000")[$P = 1000$], #math.equation(block: false, alt: "r equals 0.04")[$r = 0.04$], #math.equation(block: false, alt: "n equals 12")[$n = 12$], and #math.equation(block: false, alt: "t equals 5")[$t = 5$].
			#math.equation(block: true, alt: "A open parenthesis 5 close parenthesis, equals 1000 open parenthesis 1 plus the fraction 0.04 over 12 close parenthesis to the power 12 open parenthesis 5 close parenthesis; equals 1000 open parenthesis 1.003 close parenthesis to the power 60 equals 1221.00")[$A ( 5 ) & = 1000 attach(( 1 + frac(0.04, 12) ), t: 12 ( 5 )) \  & = 1000 ( 1.003 )^(60) = 1221.00$]
	If interest is compounded monthly, the balance in the account after #math.equation(block: false, alt: "5")[$5$] years is \$#math.equation(block: false, alt: "1221")[$1221$].
]
]


      
        Which formula gives the amount when interest is compounded quarterly?


        \_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "P open parenthesis 1 plus the fraction r over 4 close parenthesis to the power 4 t")[$P attach(( 1 + display(frac(r, 4)) ), t: 4 t)$]
]


      
      
        Which formula gives the amount when interest is compounded quarterly?


        + #math.equation(block: false, alt: "P open parenthesis 1 plus 4 r close parenthesis to the power t")[$P ( 1 + 4 r )^(t)$]
+ #math.equation(block: false, alt: "P open parenthesis 1 plus 4 r close parenthesis to the power 4 t")[$P ( 1 + 4 r )^(4 t)$]
+ #math.equation(block: false, alt: "P open parenthesis 1 plus the fraction r over 4 close parenthesis to the power 4 t")[$P attach(( 1 + display(frac(r, 4)) ), t: 4 t)$]
+ #math.equation(block: false, alt: "P open parenthesis 1 plus the fraction r over 4 close parenthesis to the power t / 4")[$P attach(( 1 + display(frac(r, 4)) ), t: t / 4)$]


      
      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
In Example, you can see that the larger the value of #math.equation(block: false, alt: "n")[$n$], the greater the value of #math.equation(block: false, alt: "A")[$A$], keeping the other parameters fixed. More frequent compounding periods result in a higher account balance.
]


      
        Calculate the amount in Rashad's account after 5 years if the interest is compounded daily. (See Example. There are 365 days in a year.)


        Answer: \$\_\_\_\_\_


        #solutionbox[
\$1221.39
]


      
      
        Calculate the amount in Rashad's account after 5 years if the interest is compounded daily. (See Example. There are 365 days in a year.)


        #solutionbox[
\$1221.39
]


      
    
    
      
=== Solving Formulas


      The techniques for solving exponential equations can also be used to solve formulas involving exponential expressions for one variable in terms of the others.


      #examplebox("Example 6")[][
Solve #math.equation(block: false, alt: "2 C equals C b to the power k t")[$" " " " 2 C = C b^(k t) " " " "$] for #math.equation(block: false, alt: "t")[$t$]. (Assume that #math.equation(block: false, alt: "C")[$C$] and #math.equation(block: false, alt: "k not equal to 0")[$k ≠ 0$].)


        
        #solutionbox[
First, we divide both sides by #math.equation(block: false, alt: "C")[$C$] to isolate the power.


          #math.equation(block: true, alt: "b to the power k t equals 2")[$b^(k t) = 2$]
          Next, we take the log base #math.equation(block: false, alt: "10")[$10$] of both sides.


          #math.equation(block: true, alt: "log ⁡ open parenthesis b to the power k t close parenthesis, equals log ⁡ 2, Apply Log Property (3).; k t log ⁡ open parenthesis b close parenthesis, equals log ⁡ 2")[$log ( b^(k t) ) & = log 2 &  & "Apply Log Property (3)." \ k t log ( b ) & = log 2$]
          Finally, we divide both sides by #math.equation(block: false, alt: "k log ⁡ open parenthesis b close parenthesis")[$k log ( b )$] to solve for #math.equation(block: false, alt: "t")[$t$].


          #math.equation(block: true, alt: "t equals the fraction log ⁡ open parenthesis 2 close parenthesis over k log ⁡ open parenthesis b close parenthesis")[$t = frac(log ( 2 ), k log ( b ))$]
]
]


      
        Solve #math.equation(block: false, alt: "A equals P open parenthesis 1 plus r close parenthesis to the power t")[$A = P ( 1 + r )^(t)$] for #math.equation(block: false, alt: "t")[$t$].


        #math.equation(block: false, alt: "t equals")[$t =$]\_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "t equals the fraction log ⁡ open parenthesis A / P close parenthesis over log ⁡ open parenthesis 1 plus r close parenthesis")[$t = display(frac(log ( A / P ), log ( 1 + r )))$]
]


      
      
        Solve #math.equation(block: false, alt: "A equals P open parenthesis 1 plus r close parenthesis to the power t")[$A = P ( 1 + r )^(t)$] for #math.equation(block: false, alt: "t")[$t$].


        #solutionbox[
#math.equation(block: true, alt: "t equals the fraction log ⁡ open parenthesis A / P close parenthesis over log ⁡ open parenthesis 1 plus r close parenthesis")[$t = display(frac(log ( A / P ), log ( 1 + r )))$]
]


      
    
    
      
=== Section Summary


      
        
==== Vocabulary


        Look up the definitions of new terms in the Glossary.


        - Compounding period


      
      
        
==== CONCEPTS


        + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Properties of Logarithms]

If #math.equation(block: false, alt: "x")[$x$], #math.equation(block: false, alt: "y")[$y$], #math.equation(block: false, alt: "b greater than 0")[$b > 0$], and #math.equation(block: false, alt: "b not equal to 1")[$b ≠ 1$], then

+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis x y close parenthesis equals log sub b ⁡ open parenthesis x close parenthesis plus log sub b ⁡ open parenthesis y close parenthesis")[$log_(b) ( x y ) = log_(b) ( x ) + log_(b) ( y )$]
+ #math.equation(block: false, alt: "log sub open parenthesis b close parenthesis ⁡ the fraction x over y equals log sub b ⁡ open parenthesis x close parenthesis minus log sub b ⁡ open parenthesis y close parenthesis")[$log_(( b )) display(frac(x, y)) = log_(b) ( x ) − log_(b) ( y )$]
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis x to the power k close parenthesis equals k log sub b ⁡ open parenthesis x close parenthesis")[$log_(b) ( x^(k) ) = k log_(b) ( x )$]
]
+ We can use the properties of logarithms to solve exponential equations with any base.#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Steps for Solving Exponential Equations]

+ Isolate the power on one side of the equation.
+ Take the log base #math.equation(block: false, alt: "10")[$10$] of both sides.
+ Simplify by applying Log Property (3).
+ Solve for the variable.
]
+ The amount in an account earning interest compounded #math.equation(block: false, alt: "n")[$n$] times per year is an exponential function of time.#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Compound Interest]

The amount #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$] accumulated (principal plus interest) in an account bearing interest compounded #math.equation(block: false, alt: "n")[$n$] times annually is

#math.equation(block: true, alt: "A open parenthesis t close parenthesis equals P open parenthesis 1 plus the fraction r over n close parenthesis to the power n t")[$A ( t ) = P attach(( 1 + frac(r, n) ), t: n t)$]where

#figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([], [], [#math.equation(block: false, alt: "P")[$P$]], [is the principal invested,]),
  [], [], [#math.equation(block: false, alt: "r")[$r$]], [is the interest rate,],
  [], [], [#math.equation(block: false, alt: "t")[$t$]], [is the time period, in years.],
))
]


      
      
        
==== STUDY QUESTIONS


        + The properties of logs are really another form of which familiar laws?
+ Which log property allows us to solve an exponential equation whose base is not #math.equation(block: false, alt: "10")[$10$]?
+ Explain why #math.equation(block: false, alt: "12 times 10 to the power 3 x")[$12 ⋅ 10^(3 x)$] is not the same as #math.equation(block: false, alt: "120 to the power 3 x")[$120^(3 x)$].
+ Which of the following expressions are equivalent?
			#math.equation(block: true, alt: "log ⁡ open parenthesis the fraction x over 4 close parenthesis the fraction log ⁡ open parenthesis x close parenthesis over log ⁡ open parenthesis 4 close parenthesis log ⁡ open parenthesis x minus 4 close parenthesis log ⁡ open parenthesis x close parenthesis minus log ⁡ open parenthesis 4 close parenthesis")[$log ( display(frac(x, 4)) ) " " " " " " " " " " display(frac(log ( x ), log ( 4 ))) " " " " " " " " " " " " " " log ( x − 4 ) " " " " " " log ( x ) − log ( 4 )$]
+ Which of the following expressions are equivalent?
			#math.equation(block: true, alt: "log ⁡ open parenthesis x plus 2 close parenthesis log ⁡ open parenthesis x close parenthesis plus log ⁡ open parenthesis 2 close parenthesis log ⁡ open parenthesis 2 x close parenthesis open parenthesis log ⁡ open parenthesis 2 close parenthesis close parenthesis open parenthesis log ⁡ open parenthesis x close parenthesis close parenthesis")[$log ( x + 2 ) " " " " " " " " " " log ( x ) + log ( 2 ) " " " " " " " " " " log ( 2 x ) " " " " " " " " " " ( log ( 2 ) ) ( log ( x ) )$]
+ Which of the following expressions are equivalent?
			#math.equation(block: true, alt: "log ⁡ open parenthesis x cubed close parenthesis open parenthesis log ⁡ open parenthesis 3 close parenthesis close parenthesis open parenthesis log ⁡ open parenthesis x close parenthesis close parenthesis 3 log ⁡ open parenthesis x close parenthesis log ⁡ open parenthesis 3 to the power x close parenthesis")[$log ( x^(3) ) " " " " " " " " " " ( log ( 3 ) ) ( log ( x ) ) " " " " " " " " " " 3 log ( x ) " " " " " " " " " " log ( 3^(x) )$]


      
      
        
==== SKILLS


        Practice each skill in the Homework problems listed.


        + Use the properties of logarithms to simplify expressions: \#1–24, \#45–52
+ Solve exponential equations using logs base 10: \#25–36
+ Solve problems about exponential models: \#37–44
+ Solve problems about compound interest: \#53–58
+ Solve formulas involving exponential expressions: \#59–64


      
    
    
      
=== Homework 4.4


      
        + Simplify #math.equation(block: false, alt: "10 squared times 10 to the power 6")[$10^(2) ⋅ 10^(6)$].
+ Compute #math.equation(block: false, alt: "log ⁡ open parenthesis 10 squared close parenthesis")[$log ( 10^(2) )$], #math.equation(block: false, alt: "log ⁡ open parenthesis 10 to the power 6 close parenthesis")[$log ( 10^(6) )$], and #math.equation(block: false, alt: "log ⁡ open parenthesis 10 squared times 10 to the power 6 close parenthesis")[$log ( 10^(2) ⋅ 10^(6) )$]. How are they related?


        
          + #math.equation(block: false, alt: "10 to the power 8")[$10^(8)$]
+ #math.equation(block: false, alt: "2")[$2$]; #math.equation(block: false, alt: "6")[$" " 6$]; #math.equation(block: false, alt: "8")[$" " 8$]; #math.equation(block: false, alt: "2 plus 6 equals 8")[$" " 2 + 6 = 8$]


        
      
      
        + Simplify #math.equation(block: false, alt: "the fraction 10 to the power 9 over 10 to the power 6")[$display(frac(10^(9), 10^(6)))$].
+ Compute #math.equation(block: false, alt: "log ⁡ open parenthesis 10 to the power 9 close parenthesis")[$log ( 10^(9) )$], #math.equation(block: false, alt: "log ⁡ open parenthesis 10 to the power 6 close parenthesis")[$log ( 10^(6) )$], and #math.equation(block: false, alt: "log ⁡ open parenthesis the fraction 10 to the power 9 over 10 to the power 6 close parenthesis")[$log ( display(frac(10^(9), 10^(6))) )$]. How are they related?


      
      
        + Simplify #math.equation(block: false, alt: "the fraction b to the power 8 over b to the power 5")[$display(frac(b^(8), b^(5)))$].
+ Compute #math.equation(block: false, alt: "log sub b ⁡ open parenthesis b to the power 8 close parenthesis")[$log_(b) ( b^(8) )$], #math.equation(block: false, alt: "log sub b ⁡ open parenthesis b to the power 5 close parenthesis")[$log_(b) ( b^(5) )$], and #math.equation(block: false, alt: "log sub b ⁡ open parenthesis the fraction b to the power 8 over b to the power 5 close parenthesis")[$log_(b) ( display(frac(b^(8), b^(5))) )$]. How are they related?


        
          + #math.equation(block: false, alt: "b cubed")[$b^(3)$]
+ #math.equation(block: false, alt: "8")[$8$]; #math.equation(block: false, alt: "5")[$" " 5$]; #math.equation(block: false, alt: "3")[$" " 3$]; #math.equation(block: false, alt: "8 minus 5 equals 3")[$" " 8 − 5 = 3$]


        
      
      
        + Simplify #math.equation(block: false, alt: "b to the power 4 times b cubed")[$b^(4) ⋅ b^(3)$].
+ Compute #math.equation(block: false, alt: "log sub b ⁡ open parenthesis b to the power 4 close parenthesis")[$log_(b) ( b^(4) )$], #math.equation(block: false, alt: "log sub b ⁡ open parenthesis b cubed close parenthesis")[$log_(b) ( b^(3) )$], and #math.equation(block: false, alt: "log sub b ⁡ open parenthesis b to the power 4 times b cubed close parenthesis")[$log_(b) ( b^(4) ⋅ b^(3) )$]. How are they related?


      
      
        + Simplify #math.equation(block: false, alt: "open parenthesis 10 cubed close parenthesis to the power 5")[$attach(( 10^(3) ), t: 5)$].
+ Compute #math.equation(block: false, alt: "log ⁡ open parenthesis 10 cubed close parenthesis to the power 5")[$log attach(( 10^(3) ), t: 5)$]  and #math.equation(block: false, alt: "log ⁡ open parenthesis 10 cubed close parenthesis")[$log ( 10^(3) )$]. How are they related?


        
          + #math.equation(block: false, alt: "10 to the power 15")[$10^(15)$]
+ #math.equation(block: false, alt: "15")[$15$]; #math.equation(block: false, alt: "3")[$" " 3$];  #math.equation(block: false, alt: "15 equals 3 times 5")[$" " 15 = 3 ⋅ 5$]


        
      
      
        + Simplify #math.equation(block: false, alt: "open parenthesis b squared close parenthesis to the power 6")[$attach(( b^(2) ), t: 6)$].
+ Compute #math.equation(block: false, alt: "log sub b ⁡ open parenthesis open parenthesis b squared close parenthesis to the power 6 close parenthesis")[$log_(b) ( attach(( b^(2) ), t: 6) )$]  and #math.equation(block: false, alt: "log sub b ⁡ open parenthesis b squared close parenthesis")[$log_(b) ( b^(2) )$]. How are they related?


      
      For Problems 7–14, use the properties of logarithms to expand each expression in terms of simpler logarithms. Assume that all variable expressions denote positive numbers.


      
        + #math.equation(block: false, alt: "log sub b ⁡ open parenthesis 2 x close parenthesis")[$log_(b) ( 2 x )$]
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis the fraction x over 2 close parenthesis")[$log_(b) ( display(frac(x, 2)) )$]


        
          + #math.equation(block: false, alt: "log sub b ⁡ open parenthesis 2 close parenthesis plus log sub b ⁡ open parenthesis x close parenthesis")[$log_(b) ( 2 ) + log_(b) ( x )$]
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis 2 close parenthesis minus log sub b ⁡ open parenthesis x close parenthesis")[$log_(b) ( 2 ) − log_(b) ( x )$]


        
      
      
        + #math.equation(block: false, alt: "log sub b ⁡ open parenthesis the fraction 2 x over x minus 2 close parenthesis")[$log_(b) ( display(frac(2 x, x − 2)) )$]
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis x open parenthesis 2 x plus 3 close parenthesis close parenthesis")[$log_(b) ( x ( 2 x + 3 ) )$]


      
      
        + #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 3 x to the power 4 close parenthesis")[$log_(3) ( 3 x^(4) )$]
+ #math.equation(block: false, alt: "log sub 5 ⁡ open parenthesis 1.1 to the power 1 / t close parenthesis")[$log_(5) ( 1.1^(1 / t) )$]


        
          + #math.equation(block: false, alt: "1 plus 4 log sub 3 ⁡ x")[$1 + 4 log_(3) x$]
+ #math.equation(block: false, alt: "the fraction 1 over t log sub 5 ⁡ open parenthesis 1.1 close parenthesis")[$display(frac(1, t)) log_(5) ( 1.1 )$]


        
      
      
        + #math.equation(block: false, alt: "log sub b ⁡ open parenthesis open parenthesis 4 b close parenthesis to the power t close parenthesis")[$log_(b) ( ( 4 b )^(t) )$]
+ #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis 5 times 2 to the power x close parenthesis")[$log_(2) ( 5 ⋅ 2^(x) )$]


      
      
        + #math.equation(block: false, alt: "log sub b ⁡ open parenthesis the square root of b x close parenthesis")[$log_(b) ( sqrt(b x) )$]
+ #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis the cube root of x squared plus 1 close parenthesis")[$log_(3) ( root(3, x^(2) + 1) )$]


        
          + #math.equation(block: false, alt: "the fraction 1 over 2 plus the fraction 1 over 2 log sub b ⁡ open parenthesis x close parenthesis")[$display(frac(1, 2)) + display(frac(1, 2)) log_(b) ( x )$]
+ #math.equation(block: false, alt: "the fraction 1 over 3 log sub 3 ⁡ open parenthesis x squared plus 1 close parenthesis")[$display(frac(1, 3)) log_(3) ( x^(2) + 1 )$]


        
      
      
        + #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis the square root of the fraction 2 L over R squared close parenthesis")[$log_(10) ( sqrt(display(frac(2 L, R^(2)))) )$]
+ #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 2 π the square root of the fraction l over g close parenthesis")[$log_(10) ( 2 π sqrt(display(frac(l, g))) )$]


      
      
        + #math.equation(block: false, alt: "log ⁡ open parenthesis P sub 0 open parenthesis 1 minus m close parenthesis to the power t close parenthesis")[$log ( P_(0) ( 1 − m )^(t) )$]
+ #math.equation(block: false, alt: "log sub 4 ⁡ open parenthesis 1 plus the fraction r over 4 close parenthesis to the power 4 t")[$log_(4) attach(( 1 + display(frac(r, 4)) ), t: 4 t)$]


        
          + #math.equation(block: false, alt: "log ⁡ P sub 0 plus t log ⁡ open parenthesis 1 minus m close parenthesis")[$log P_(0) + t log ( 1 − m )$]
+ #math.equation(block: false, alt: "4 t open bracket log sub 4 ⁡ open parenthesis 4 plus r close parenthesis minus 1 close bracket")[$4 t [ log_(4) ( 4 + r ) − 1 ]$]


        
      
      
        + #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis the fraction a squared minus 2 over a to the power 5 close parenthesis")[$log_(3) ( display(frac(a^(2) − 2, a^(5))) )$]
+ #math.equation(block: false, alt: "log ⁡ open parenthesis the fraction a cubed b squared over open parenthesis a plus b close parenthesis to the power 3 / 2 close parenthesis")[$log ( display(frac(a^(3) b^(2), ( a + b )^(3 / 2))) )$]


      
      For Problems 15–20, combine into one logarithm and simplify. Assume all expressions are defined.


      
        + #math.equation(block: false, alt: "log sub b ⁡ open parenthesis 8 close parenthesis minus log sub b ⁡ open parenthesis 2 close parenthesis")[$log_(b) ( 8 ) − log_(b) ( 2 )$]
+ #math.equation(block: false, alt: "2 log sub 4 ⁡ open parenthesis x close parenthesis plus 3 log sub 4 ⁡ open parenthesis y close parenthesis")[$2 log_(4) ( x ) + 3 log_(4) ( y )$]


        
          + #math.equation(block: false, alt: "log sub b ⁡ open parenthesis 4 close parenthesis")[$log_(b) ( 4 )$]
+ #math.equation(block: false, alt: "log sub 4 ⁡ open parenthesis x squared y cubed close parenthesis")[$log_(4) ( x^(2) y^(3) )$]


        
      
      
        + #math.equation(block: false, alt: "log sub b ⁡ open parenthesis 5 close parenthesis plus log sub b ⁡ open parenthesis 2 close parenthesis")[$log_(b) ( 5 ) + log_(b) ( 2 )$]
+ #math.equation(block: false, alt: "the fraction 1 over 4 log sub 5 ⁡ open parenthesis x close parenthesis minus the fraction 3 over 4 log sub 5 ⁡ open parenthesis y close parenthesis")[$display(frac(1, 4)) log_(5) ( x ) − display(frac(3, 4)) log_(5) ( y )$]


      
      
        + #math.equation(block: false, alt: "log ⁡ open parenthesis 2 x close parenthesis plus 2 log ⁡ open parenthesis x close parenthesis minus log ⁡ open parenthesis the square root of x close parenthesis")[$log ( 2 x ) + 2 log ( x ) − log ( sqrt(x) )$]
+ #math.equation(block: false, alt: "log ⁡ open parenthesis t squared minus 16 close parenthesis minus log ⁡ open parenthesis t plus 4 close parenthesis")[$log ( t^(2) − 16 ) − log ( t + 4 )$]


        
          + #math.equation(block: false, alt: "log ⁡ open parenthesis 2 x to the power 5 / 2 close parenthesis")[$log ( 2 x^(5 / 2) )$]
+ #math.equation(block: false, alt: "log ⁡ open parenthesis t minus 4 close parenthesis")[$log ( t − 4 )$]


        
      
      
        + #math.equation(block: false, alt: "log ⁡ open parenthesis x squared close parenthesis plus log ⁡ open parenthesis x cubed close parenthesis minus 5 log ⁡ open parenthesis x close parenthesis")[$log ( x^(2) ) + log ( x^(3) ) − 5 log ( x )$]
+ #math.equation(block: false, alt: "log ⁡ open parenthesis x squared minus x close parenthesis minus log ⁡ open parenthesis the square root of x cubed close parenthesis")[$log ( x^(2) − x ) − log ( sqrt(x^(3)) )$]


      
      
        + #math.equation(block: false, alt: "3 minus 3 log ⁡ open parenthesis 30 close parenthesis")[$3 − 3 log ( 30 )$]
+ #math.equation(block: false, alt: "the fraction 1 over 3 log sub 6 ⁡ open parenthesis 8 w to the power 6 close parenthesis")[$display(frac(1, 3)) log_(6) ( 8 w^(6) )$]


        
          + #math.equation(block: false, alt: "log ⁡ the fraction 1 over 27")[$log display(frac(1, 27))$]
+ #math.equation(block: false, alt: "log sub 6 ⁡ open parenthesis 2 w squared close parenthesis")[$log_(6) ( 2 w^(2) )$]


        
      
      
        + #math.equation(block: false, alt: "2 minus log sub 4 ⁡ open parenthesis 16 z squared close parenthesis")[$2 − log_(4) ( 16 z^(2) )$]
+ #math.equation(block: false, alt: "1 minus 2 log sub 3 ⁡ open parenthesis x close parenthesis")[$1 − 2 log_(3) ( x )$]


      
      For Problems 21–24, use the three logs below to find the value of each expression.


      #math.equation(block: true, alt: "log sub b ⁡ open parenthesis 2 close parenthesis equals 0.6931 , log sub b ⁡ open parenthesis 3 close parenthesis equals 1.0986 , log sub b ⁡ open parenthesis 5 close parenthesis equals 1.6094")[$log_(b) ( 2 ) = 0.6931 , " " " " log_(b) ( 3 ) = 1.0986 , " " " " log_(b) ( 5 ) = 1.6094$]
      (#emph[Hint]: For example, #math.equation(block: false, alt: "log sub b ⁡ open parenthesis 15 close parenthesis equals log sub b ⁡ open parenthesis 3 close parenthesis plus log sub b ⁡ open parenthesis 5 close parenthesis")[$log_(b) ( 15 ) = log_(b) ( 3 ) + log_(b) ( 5 )$].)


      
        + #math.equation(block: false, alt: "log sub b ⁡ open parenthesis 6 close parenthesis")[$log_(b) ( 6 )$]
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis the fraction 2 over 5 close parenthesis")[$log_(b) ( display(frac(2, 5)) )$]


        
          + #math.equation(block: false, alt: "1.7917")[$1.7917$]
+ #math.equation(block: false, alt: "minus 0.9163")[$− 0.9163$]


        
      
      
        + #math.equation(block: false, alt: "log sub b ⁡ open parenthesis 10 close parenthesis")[$log_(b) ( 10 )$]
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis the fraction 3 over 2 close parenthesis")[$log_(b) ( display(frac(3, 2)) )$]


      
      
        + #math.equation(block: false, alt: "log sub b ⁡ open parenthesis 9 close parenthesis")[$log_(b) ( 9 )$]
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis the square root of 50 close parenthesis")[$log_(b) ( sqrt(50) )$]


        
          + #math.equation(block: false, alt: "2.1972")[$2.1972$]
+ #math.equation(block: false, alt: "1.9560")[$1.9560$]


        
      
      
        + #math.equation(block: false, alt: "log sub b ⁡ open parenthesis 25 close parenthesis")[$log_(b) ( 25 )$]
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis 75 close parenthesis")[$log_(b) ( 75 )$]


      
      For Problems 25–36, solve the equation by using logarithms base #math.equation(block: false, alt: "10")[$10$]. Round answers to four decimal places.


      
        #math.equation(block: true, alt: "2 to the power x equals 7")[$2^(x) = 7$]


        
          #math.equation(block: true, alt: "2.8074")[$2.8074$]


        
      
      
        #math.equation(block: true, alt: "3 to the power x equals 4")[$3^(x) = 4$]


      
      
        #math.equation(block: true, alt: "3 to the power x plus 1 equals 8")[$3^(x + 1) = 8$]


        
          #math.equation(block: true, alt: "0.8928")[$0.8928$]


        
      
      
        #math.equation(block: true, alt: "2 to the power x minus 1 equals 9")[$2^(x − 1) = 9$]


      
      
        #math.equation(block: true, alt: "4 to the power x squared equals 15")[$4^(x^(2)) = 15$]


        
          #math.equation(block: true, alt: "plus or minus 1.3977")[$± 1.3977$]


        
      
      
        #math.equation(block: true, alt: "3 to the power x squared equals 21")[$3^(x^(2)) = 21$]


      
      
        #math.equation(block: true, alt: "4.26 to the power minus x equals 10.3")[$4.26^(− x) = 10.3$]


        
          #math.equation(block: true, alt: "minus 1.6092")[$− 1.6092$]


        
      
      
        #math.equation(block: true, alt: "2.13 to the power minus x equals 8.1")[$2.13^(− x) = 8.1$]


      
      
        #math.equation(block: true, alt: "25 times 3 to the power 2.1 x equals 47")[$25 ⋅ 3^(2.1 x) = 47$]


        
          #math.equation(block: true, alt: "0.2736")[$0.2736$]


        
      
      
        #math.equation(block: true, alt: "12 times 5 to the power 1.5 x equals 85")[$12 ⋅ 5^(1.5 x) = 85$]


      
      
        #math.equation(block: true, alt: "3600 equals 20 times 8 to the power minus 0.2 x")[$3600 = 20 ⋅ 8^(− 0.2 x)$]


        
          #math.equation(block: true, alt: "minus 12.4864")[$− 12.4864$]


        
      
      
        #math.equation(block: true, alt: "0.06 equals 50 times 4 to the power minus 0.6 x")[$0.06 = 50 ⋅ 4^(− 0.6 x)$]


      
      
        If raw meat is allowed to thaw at #math.equation(block: false, alt: "50 degrees")[$50^(∘)$]F, Salmonella grows at a rate of #math.equation(block: false, alt: "9 %")[$9 upright(%)$] per hour.


        + Write a formula for the amount of Salmonella present after #math.equation(block: false, alt: "t")[$t$] hours, if the initial amount is #math.equation(block: false, alt: "S sub 0")[$S_(0)$].
+ Health officials advise that the amount of Salmonella initially present in meat should not be allowed to increase by more than #math.equation(block: false, alt: "50 %")[$50 upright(%)$]. How long can meat be left to thaw at #math.equation(block: false, alt: "50 degrees")[$50^(∘)$]F?


        
          + #math.equation(block: false, alt: "S open parenthesis t close parenthesis equals S sub 0 open parenthesis 1.09 close parenthesis to the power t")[$S ( t ) = S_(0) ( 1.09 )^(t)$]
+ #math.equation(block: false, alt: "4.7")[$4.7$] hours


        
      
      
        Starting in 1998, the demand for electricity in Ireland grew at a rate of #math.equation(block: false, alt: "5.8 %")[$5.8 upright(%)$] per year. In 1998, #math.equation(block: false, alt: "20 , 500")[$20 , 500$] gigawatts were used. (Source: Electricity Supply Board of Ireland)


        + Write a formula for electricity demand in Ireland as a function of time.
+ If demand continues to grow at the same rate, when would it reach #math.equation(block: false, alt: "30 , 000")[$30 , 000$] gigawatts?


      
      
        The concentration of a certain drug injected into the bloodstream decreases by #math.equation(block: false, alt: "20 %")[$20 upright(%)$] each hour as the drug is eliminated from the body. The initial dose creates a concentration of #math.equation(block: false, alt: "0.7")[$0.7$] milligrams per milliliter.


        + Write a function for the concentration of the drug as a function of time.
+ The minimum effective concentration of the drug is #math.equation(block: false, alt: "0.4")[$0.4$] milligrams per milliliter. When should the second dose be administered?
+ Verify your answer with a graph.


        
          + #math.equation(block: false, alt: "C open parenthesis t close parenthesis equals 0.7 open parenthesis 0.80 close parenthesis to the power t")[$C ( t ) = 0.7 ( 0.80 )^(t)$]
+ After #math.equation(block: false, alt: "2.5")[$2.5$] hours
+ #figure(figph[decay], alt: "decay", caption: none)


        
      
      
        A small pond is tested for pollution, and the concentration of toxic chemicals is found to be #math.equation(block: false, alt: "80")[$80$] parts per million. Clean water enters the pond from a stream, mixes with the polluted water, then leaves the pond so that the pollution level is reduced by #math.equation(block: false, alt: "10 %")[$10 upright(%)$] each month.


        + Write a function for the concentration of toxic chemicals as a function of time.
+ How long will it be before the concentration of toxic chemicals reaches a safe level of #math.equation(block: false, alt: "25")[$25$] parts per million?
+ Verify your answer with a graph.


      
      
        According to the National Council of Churches, the fastest growing denomination in the United States in #math.equation(block: false, alt: "2004")[$2004$] was the Jehovah's Witnesses, with an annual growth rate of #math.equation(block: false, alt: "1.82 %")[$1.82 upright(%)$].


        + The Jehovah's Witnesses had #math.equation(block: false, alt: "1 , 041 , 000")[$1 , 041 , 000$] members in #math.equation(block: false, alt: "2004")[$2004$]. Write a formula for the membership in the Jehovah's Witnesses as a function of time, assuming that the church continues to grow at the same rate.
+ When will the Jehovah's Witnesses have #math.equation(block: false, alt: "2 , 000 , 000")[$2 , 000 , 000$] members?


        
          + #math.equation(block: false, alt: "J open parenthesis t close parenthesis equals 1 , 041 , 000 times 1.0182 to the power t")[$J ( t ) = 1 , 041 , 000 ⋅ 1.0182^(t)$]
+ In #math.equation(block: false, alt: "2040")[$2040$]


        
      
      
        In #math.equation(block: false, alt: "2004")[$2004$], the Presbyterian Church had #math.equation(block: false, alt: "3 , 241 , 000")[$3 , 241 , 000$] members, but membership was declining by #math.equation(block: false, alt: "4.87 %")[$4.87 upright(%)$] annually.


        + Write a formula for the membership in the Presbyterian Church as a function of time, assuming that the membership continues to decline at the same rate.
+ When will the Presbyterian Church have #math.equation(block: false, alt: "2 , 000 , 000")[$2 , 000 , 000$] members?


      
      
        Sodium-24 is a radioactive isotope that is used in diagnosing circulatory disease. It decays into stable isotopes of sodium at a rate of #math.equation(block: false, alt: "4.73 %")[$4.73 upright(%)$] per hour.


        + Technicians inject a quantity of sodium-24 into a patient's bloodstream. Write a formula for the amount of sodium-24 present in the bloodstream as a function of time.
+ How long will it take for #math.equation(block: false, alt: "75 %")[$75 upright(%)$] of the isotope to decay?


        
          + #math.equation(block: false, alt: "S open parenthesis t close parenthesis equals S sub 0 times 0.9527 to the power t")[$S ( t ) = S_(0) ⋅ 0.9527^(t)$]
+ #math.equation(block: false, alt: "28.61")[$28.61$] hours


        
      
      
        The population of Afghanistan is growing at #math.equation(block: false, alt: "2.6 %")[$2.6 upright(%)$] per year.


        + Write a formula for the population of Afghanistan as a function of time.
+ In 2005, the population of Afghanistan was #math.equation(block: false, alt: "29.9")[$29.9$] million. At the given rate of growth, how long would it take the population to reach #math.equation(block: false, alt: "40")[$40$] million?


      
      For Problems 46–52, evaluate each expression. Which (if any) are equal?


      
        + #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis 4 times 8 close parenthesis")[$log_(2) ( 4 ⋅ 8 )$]
+ #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis 4 close parenthesis times log sub 2 ⁡ open parenthesis 8 close parenthesis close parenthesis")[$log_(2) \( 4 \) ⋅ log_(2) \( 8 \) \)$]
+ #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis 4 close parenthesis plus log sub 2 ⁡ open parenthesis 8 close parenthesis")[$log_(2) ( 4 ) + log_(2) ( 8 )$]


        
          + #math.equation(block: false, alt: "5")[$5$]
+ #math.equation(block: false, alt: "6")[$6$]
+ #math.equation(block: false, alt: "5")[$5$]


          (a) and (c) are equal.


        
      
      
        + #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis 16 plus 16 close parenthesis")[$log_(2) ( 16 + 16 )$]
+ #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis 16 close parenthesis plus log sub 2 ⁡ open parenthesis 16 close parenthesis")[$log_(2) ( 16 ) + log_(2) ( 16 )$]
+ #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis 2 close parenthesis plus log sub 2 ⁡ open parenthesis 16 close parenthesis")[$log_(2) ( 2 ) + log_(2) ( 16 )$]


      
      
        + #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 27 squared close parenthesis")[$log_(3) ( 27^(2) )$]
+ #math.equation(block: false, alt: "open parenthesis log sub 3 ⁡ open parenthesis 27 close parenthesis close parenthesis squared")[$( log_(3) ( 27 ) )^(2)$]
+ #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 27 close parenthesis plus log sub 3 ⁡ open parenthesis 27 close parenthesis")[$log_(3) ( 27 ) + log_(3) ( 27 )$]


        
          + #math.equation(block: false, alt: "6")[$6$]
+ #math.equation(block: false, alt: "9")[$9$]
+ #math.equation(block: false, alt: "6")[$6$]


          (a) and (c) are equal.


        
      
      
        + #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 3 times 27 close parenthesis")[$log_(3) ( 3 ⋅ 27 )$]
+ #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 3 close parenthesis plus log sub 3 ⁡ open parenthesis 27 close parenthesis")[$log_(3) ( 3 ) + log_(3) ( 27 )$]
+ #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 3 close parenthesis times log sub 3 ⁡ open parenthesis 27 close parenthesis")[$log_(3) ( 3 ) ⋅ log_(3) ( 27 )$]


      
      
        + #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis the fraction 240 over 10 close parenthesis")[$log_(10) ( display(frac(240, 10)) )$]
+ #math.equation(block: false, alt: "the fraction log sub 10 ⁡ open parenthesis 240 close parenthesis over log sub 10 ⁡ open parenthesis 10 close parenthesis")[$display(frac(log_(10) ( 240 ), log_(10) ( 10 )))$]
+ #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 240 minus 10 close parenthesis")[$log_(10) ( 240 − 10 )$]


        
          + #math.equation(block: false, alt: "log ⁡ 24 approximately equals 1.38")[$log 24 ≈ 1.38$]
+ #math.equation(block: false, alt: "log ⁡ 240 approximately equals 2.38")[$log 240 ≈ 2.38$]
+ #math.equation(block: false, alt: "log ⁡ 230 approximately equals 2.36")[$log 230 ≈ 2.36$]


          None are equal.


        
      
      
        + #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis the fraction 1 over 2 times 80 close parenthesis")[$log_(10) ( display(frac(1, 2)) ⋅ 80 )$]
+ #math.equation(block: false, alt: "the fraction 1 over 2 log sub 10 ⁡ open parenthesis 80 close parenthesis")[$display(frac(1, 2)) log_(10) ( 80 )$]
+ #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis the square root of 80 close parenthesis")[$log_(10) ( sqrt(80) )$]


      
      
        + #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 75 minus 15 close parenthesis")[$log_(10) ( 75 − 15 )$]
+ #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 75 close parenthesis minus log sub 10 ⁡ open parenthesis 15 close parenthesis")[$log_(10) ( 75 ) − log_(10) ( 15 )$]
+ #math.equation(block: false, alt: "the fraction log sub 10 ⁡ open parenthesis 75 close parenthesis over log sub 10 ⁡ open parenthesis 15 close parenthesis")[$display(frac(log_(10) ( 75 ), log_(10) ( 15 )))$]


        
          + #math.equation(block: false, alt: "log ⁡ open parenthesis 60 close parenthesis approximately equals 1.78")[$log ( 60 ) ≈ 1.78$]
+ #math.equation(block: false, alt: "log ⁡ open parenthesis 5 close parenthesis approximately equals 0.70")[$log ( 5 ) ≈ 0.70$]
+ #math.equation(block: false, alt: "the fraction log sub 10 ⁡ open parenthesis 75 close parenthesis over log sub 10 ⁡ open parenthesis 15 close parenthesis approximately equals 1.59")[$display(frac(log_(10) ( 75 ), log_(10) ( 15 ))) ≈ 1.59$]


          None are equal.


        
      
      
        + #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 8 times 25 close parenthesis")[$log_(10) ( 8 ⋅ 25 )$]
+ #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 25 to the power 8 close parenthesis")[$log_(10) ( 25^(8) )$]
+ #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 8 close parenthesis plus 25")[$log_(10) ( 8 ) + 25$]


      
      For Problems 53–58, use the formula for compound interest,


      #math.equation(block: true, alt: "A equals P open parenthesis 1 plus the fraction r over n close parenthesis to the power n t")[$A = P attach(( 1 + display(frac(r, n)) ), t: n t)$]
      
        What rate of interest is required so that #math.equation(block: false, alt: "$ 1000")[$upright(\$) 1000$] will yield #math.equation(block: false, alt: "$ 1900")[$upright(\$) 1900$] after #math.equation(block: false, alt: "5")[$5$] years if the interest rate is compounded monthly?


        
          #math.equation(block: true, alt: "12.9 %")[$12.9 upright(%)$]


        
      
      
        What rate of interest is required so that #math.equation(block: false, alt: "$ 400")[$upright(\$) 400$] will yield #math.equation(block: false, alt: "$ 600")[$upright(\$) 600$] after #math.equation(block: false, alt: "3")[$3$] years if the interest rate is compounded quarterly?


      
      
        How long will it take a sum of money to triple if it is invested at #math.equation(block: false, alt: "10 %")[$10 upright(%)$] compounded daily?


        
          About #math.equation(block: false, alt: "11")[$11$] years


        
      
      
        How long will it take a sum of money to increase by a factor of #math.equation(block: false, alt: "5")[$5$] if it is invested at #math.equation(block: false, alt: "10 %")[$10 upright(%)$] compounded quarterly?


      
      
        + Suppose you invest #math.equation(block: false, alt: "$ 1000")[$upright(\$) 1000$] at #math.equation(block: false, alt: "12 %")[$12 upright(%)$] annual interest for #math.equation(block: false, alt: "5")[$5$] years. In this problem, we will investigate how the number of compounding periods,#math.equation(block: false, alt: "n")[$n$], affects the amount, #math.equation(block: false, alt: "A")[$A$]. Write #math.equation(block: false, alt: "A")[$A$] as a function of #math.equation(block: false, alt: "n")[$n$], with #math.equation(block: false, alt: "P equals 1000")[$P = 1000$], #math.equation(block: false, alt: "r equals 0.12")[$r = 0.12$], and #math.equation(block: false, alt: "t equals 5")[$t = 5$].
+ Use your calculator to make a table of values for #math.equation(block: false, alt: "A")[$A$] as a function of #math.equation(block: false, alt: "n")[$n$]. What happens to #math.equation(block: false, alt: "A")[$A$] as #math.equation(block: false, alt: "n")[$n$] increases?
+ What value of #math.equation(block: false, alt: "n")[$n$] is necessary to produce an amount #math.equation(block: false, alt: "A greater than 1818")[$A > 1818$]? To produce #math.equation(block: false, alt: "A greater than 1820")[$A > 1820$]? To produce #math.equation(block: false, alt: "A greater than 1822")[$A > 1822$]?
+ Graph the function #math.equation(block: false, alt: "A open parenthesis n close parenthesis")[$A ( n )$] in the window
       #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 52; Ymin, equals 1750, Ymax equals 1850")[$"Xmin" & = 0 &  & "Xmax" = 52 \ "Ymin" & = 1750 &  & "Ymax" = 1850$]
        Describe the graph: Is it increasing or decreasing? Concave up or down? Does it appear to have an asymptote? Give your best estimate for the asymptote.


        
          + #math.equation(block: false, alt: "A equals 1000 open parenthesis 1 plus the fraction 0.12 over n close parenthesis to the power 5 n")[$A = 1000 attach(( 1 + display(frac(0.12, n)) ), t: 5 n)$]
+ #figure(figph[table], alt: "table", caption: none)

 #linebreak() #math.equation(block: false, alt: "A")[$A$] increases.
+ #math.equation(block: false, alt: "16")[$16$]; #math.equation(block: false, alt: "31")[$31$]; #math.equation(block: false, alt: "553")[$553$]
+ Increasing, concave down, asymptotically approaching #math.equation(block: false, alt: "A approximately equals 1822.12")[$A ≈ 1822.12$]


        
      
      
        + In this problem we will repeat Problem 57 for #math.equation(block: false, alt: "4 %")[$4 upright(%)$] interest. Write #math.equation(block: false, alt: "A")[$A$] as a function of #math.equation(block: false, alt: "n")[$n$], with #math.equation(block: false, alt: "P equals 1000")[$P = 1000$], #math.equation(block: false, alt: "r equals 0.04")[$r = 0.04$], and #math.equation(block: false, alt: "t equals 5")[$t = 5$].
+ Use your calculator to make a table of values for #math.equation(block: false, alt: "A")[$A$] as a function of #math.equation(block: false, alt: "n")[$n$]. What happens to #math.equation(block: false, alt: "A")[$A$] as #math.equation(block: false, alt: "n")[$n$] increases?
+ What value of #math.equation(block: false, alt: "n")[$n$] is necessary to produce an amount #math.equation(block: false, alt: "A greater than 1218")[$A > 1218$]? To produce #math.equation(block: false, alt: "A greater than 1220")[$A > 1220$]? To produce #math.equation(block: false, alt: "A greater than 1221.40")[$A > 1221.40$]?
+ Graph the function #math.equation(block: false, alt: "A open parenthesis n close parenthesis")[$A ( n )$] in the window
       #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 52; Ymin, equals 1210, Ymax equals 1225")[$"Xmin" & = 0 &  & "Xmax" = 52 \ "Ymin" & = 1210 &  & "Ymax" = 1225$]
        Describe the graph: Is it increasing or decreasing? Concave up or down? Does it appear to have an asymptote? Give your best estimate for the asymptote.


      
      For Problems 59–64, solve the formula for the specified variable.


      
        #math.equation(block: false, alt: "N equals N sub 0 a to the power k t")[$N = N_(0) a^(k t)$], for #math.equation(block: false, alt: "k")[$k$]


        
          #math.equation(block: true, alt: "k equals the fraction 1 over t the fraction log ⁡ open parenthesis N / N sub 0 close parenthesis over log ⁡ open parenthesis a close parenthesis")[$k = display(frac(1, t)) display(frac(log ( N / N_(0) ), log ( a )))$]


        
      
      
        #math.equation(block: false, alt: "Q equals Q sub 0 b to the power t / 2")[$Q = Q_(0) b^(t / 2)$], for #math.equation(block: false, alt: "t")[$t$]


      
      
        #math.equation(block: false, alt: "A equals A sub 0 open parenthesis 10 to the power k t minus 1 close parenthesis")[$A = A_(0) ( 10^(k t) − 1 )$], for #math.equation(block: false, alt: "t")[$t$]


        
          #math.equation(block: true, alt: "t equals the fraction 1 over k log ⁡ open parenthesis the fraction A over A sub 0 plus 1 close parenthesis")[$t = display(frac(1, k)) log ( display(frac(A, A_(0))) + 1 )$]


        
      
      
        #math.equation(block: false, alt: "B equals B sub 0 open parenthesis 1 minus 10 to the power minus k t close parenthesis")[$B = B_(0) ( 1 − 10^(− k t) )$], for #math.equation(block: false, alt: "t")[$t$]


      
      
        #math.equation(block: false, alt: "w equals p v to the power q")[$w = p v^(q)$], for #math.equation(block: false, alt: "q")[$q$]


        
          #math.equation(block: true, alt: "q equals the fraction log ⁡ open parenthesis w / p close parenthesis over log ⁡ open parenthesis v close parenthesis")[$q = display(frac(log ( w / p ), log ( v )))$]


        
      
      
        #math.equation(block: false, alt: "l equals p to the power a q to the power b")[$l = p^(a) q^(b)$], for #math.equation(block: false, alt: "b")[$b$]


      
      In Problems 65–68 we use the laws of exponents to prove the properties of logarithms.


      
        We will use the first law of exponents, #math.equation(block: false, alt: "a to the power p times a to the power q equals a to the power p plus q")[$a^(p) ⋅ a^(q) = a^(p + q)$], to prove the first property of logarithms.


        + Let #math.equation(block: false, alt: "m equals log sub b ⁡ open parenthesis x close parenthesis")[$m = log_(b) ( x )$] and #math.equation(block: false, alt: "n equals log sub b ⁡ open parenthesis y close parenthesis")[$n = log_(b) ( y )$]. Rewrite these equations in exponential form:
        #math.equation(block: true, alt: "x equals bar and y equals bar")[$x = underline(#h(4em)) " " " " " " " and " " " " " " " y = underline(#h(4em))$]
+ Now consider the expression #math.equation(block: false, alt: "log sub b ⁡ open parenthesis x y close parenthesis")[$log_(b) ( x y )$]. Replace #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] by your answers to part (a).
+ Apply the first law of exponents to your expression in part (b).
+ Use the definition of logarithm to simplify your answer to part (c).
+ Refer to the definitions of #math.equation(block: false, alt: "m")[$m$] and #math.equation(block: false, alt: "n")[$n$] in part (a) to finish the proof.


        
          + #math.equation(block: false, alt: "x equals b to the power m")[$x = b^(m)$], #math.equation(block: false, alt: "y equals b to the power n")[$y = b^(n)$]
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis b to the power m times b to the power n close parenthesis")[$log_(b) ( b^(m) ⋅ b^(n) )$]
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis b to the power m times b to the power n close parenthesis equals log sub b ⁡ open parenthesis b to the power m plus n close parenthesis")[$log_(b) ( b^(m) ⋅ b^(n) ) = log_(b) ( b^(m + n) )$]
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis b to the power m plus n close parenthesis equals m plus n")[$log_(b) ( b^(m + n) ) = m + n$]
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis b to the power m plus n close parenthesis equals open parenthesis log sub b ⁡ open parenthesis x close parenthesis close parenthesis plus open parenthesis log sub b ⁡ open parenthesis y close parenthesis close parenthesis")[$log_(b) ( b^(m + n) ) = ( log_(b) ( x ) ) + ( log_(b) ( y ) )$]


        
      
      
        We will use the second law of exponents, #math.equation(block: false, alt: "the fraction a to the power p over a to the power q equals a to the power p minus q")[$display(frac(a^(p), a^(q))) = a^(p − q)$], to prove the second property of logarithms.


        + Let #math.equation(block: false, alt: "m equals log sub b ⁡ open parenthesis x close parenthesis")[$m = log_(b) ( x )$] and #math.equation(block: false, alt: "n equals log sub b ⁡ open parenthesis y close parenthesis")[$n = log_(b) ( y )$]. Rewrite these equations in exponential form:
        #math.equation(block: true, alt: "x equals bar and y equals bar")[$x = underline(#h(4em)) " " " " " " " and " " " " " " " y = underline(#h(4em))$]
+ Now consider the expression #math.equation(block: false, alt: "log sub b ⁡ open parenthesis the fraction x over y close parenthesis")[$log_(b) ( frac(x, y) )$]. Replace #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] by your answers to part (a).
+ Apply the second law of exponents to your expression in part (b).
+ Use the definition of logarithm to simplify your answer to part (c).
+ Refer to the definitions of #math.equation(block: false, alt: "m")[$m$] and #math.equation(block: false, alt: "n")[$n$] in part (a) to finish the proof.


      
      
        We will use the third law of exponents, #math.equation(block: false, alt: "open parenthesis a to the power p close parenthesis to the power q equals a to the power p q")[$( a^(p) )^(q) = a^(p q)$], to prove the third property of logarithms.


        + Let #math.equation(block: false, alt: "m equals log sub b ⁡ open parenthesis x close parenthesis")[$m = log_(b) ( x )$]. Rewrite this equation in exponential form:
        #math.equation(block: true, alt: "x equals bar")[$x = underline(#h(7.5em))$]
+ Now consider the expression #math.equation(block: false, alt: "log sub b ⁡ open parenthesis x to the power k close parenthesis")[$log_(b) ( x^(k) )$]. Replace #math.equation(block: false, alt: "x")[$x$] by your answers to part (a).
+ Apply the third law of exponents to your expression in part (b).
+ Use the definition of logarithm to simplify your answer to part (c).
+ Refer to the definitions of #math.equation(block: false, alt: "m")[$m$] in part (a) to finish the proof.


        
          + #math.equation(block: false, alt: "x equals b to the power m")[$x = b^(m)$]
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis open parenthesis b to the power m close parenthesis to the power k close parenthesis")[$log_(b) ( ( b^(m) )^(k) )$]
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis open parenthesis b to the power m close parenthesis to the power k close parenthesis equals log sub b ⁡ open parenthesis b to the power m k close parenthesis")[$log_(b) ( ( b^(m) )^(k) ) = log_(b) ( b^(m k) )$]
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis b to the power m k close parenthesis equals m k")[$log_(b) ( b^(m k) ) = m k$]
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis b to the power m k close parenthesis equals open parenthesis log sub b ⁡ open parenthesis x close parenthesis close parenthesis times k")[$log_(b) ( b^(m k) ) = ( log_(b) ( x ) ) ⋅ k$]


        
      
      
        + Use the logarithms and exponents conversion equations to explain why #math.equation(block: false, alt: "log sub b ⁡ open parenthesis 1 close parenthesis equals 0")[$log_(b) ( 1 ) = 0$].
+ Use the logarithms and exponents conversion equations  to explain why #math.equation(block: false, alt: "log sub b ⁡ open parenthesis b to the power x close parenthesis equals x")[$log_(b) ( b^(x) ) = x$].
+ Use the logarithms and exponents conversion equations  to explain why #math.equation(block: false, alt: "b to the power log sub b ⁡ open parenthesis x close parenthesis equals x")[$b^(log_(b) ( x )) = x$].