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== 10.3#h(0.6em)Evaluate and Graph Logarithmic Functions

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Before you get started, take this readiness quiz.

Solve: #math.equation(block: false, alt: "x squared equals 81 .")[$x^(2) = 81 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "x equals 9 , x equals minus 9")[$x = 9 , #h(0.5em) x = − 9$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Evaluate: #math.equation(block: false, alt: "3 to the power −2 .")[$3^(−2) .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "the fraction 1 over 9")[$frac(1, 9)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "2 to the power 4 equals 3 x minus 5 .")[$2^(4) = 3 x − 5 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "x equals 7")[$x = 7$]
]
]

We have spent some time finding the inverse of many functions. It works well to ‘undo’ an operation with another operation. Subtracting ‘undoes’ addition, multiplication ‘undoes’ division, taking the square root ‘undoes’ squaring.

As we studied the exponential function, we saw that it is one-to-one as its graphs pass the horizontal line test. This means an exponential function does have an inverse. If we try our algebraic method for finding an inverse, we run into a problem.

#math.equation(block: true, alt: "Rewrite with y equals f open parenthesis x close parenthesis .; Interchange the variables x and y ., f open parenthesis x close parenthesis, equals, a to the power x; y, equals, a to the power x; x, equals, a to the power y; Solve for y ., Oops! We have no way to solve for y !")[$"Rewrite with" #h(0.2em) y = f ( x ) . \ "Interchange the variables" #h(0.2em) x #h(0.2em) "and" #h(0.2em) y . &  &  & f ( x ) & = & a^(x) \ y & = & a^(x) \ x & = & a^(y) \ "Solve for" #h(0.2em) y . &  &  & "Oops! We have no way to solve for" #h(0.2em) y !$]

To deal with this we define the logarithm function with base #emph[a] to be the inverse of the exponential function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a to the power x .")[$f ( x ) = a^(x) .$] We use the notation #math.equation(block: false, alt: "f to the power −1 open parenthesis x close parenthesis equals log sub a x")[$f^(−1) ( x ) = "log"_(a) x$] and say the inverse function of the exponential function is the logarithmic function.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Logarithmic Function]

The function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals log sub a x")[$f ( x ) = "log"_(a) x$] is the #strong[logarithmic function] with base #math.equation(block: false, alt: "a")[$a$], where #math.equation(block: false, alt: "a greater than 0 ,")[$a > 0 ,$]#math.equation(block: false, alt: "x greater than 0 ,")[$x > 0 ,$] and #math.equation(block: false, alt: "a not equal to 1 .")[$a ≠ 1 .$]

#math.equation(block: true, alt: "y equals log sub a x is equivalent to x equals a to the power y")[$y = "log"_(a) x #h(0.2em) "is equivalent to" #h(0.2em) x = a^(y)$]
]


=== Convert Between Exponential and Logarithmic Form

Since the equations #math.equation(block: false, alt: "y equals log sub a x")[$y = "log"_(a) x$] and #math.equation(block: false, alt: "x equals a to the power y")[$x = a^(y)$] are equivalent, we can go back and forth between them. This will often be the method to solve some exponential and logarithmic equations. To help with converting back and forth let’s take a close look at the equations. Notice the positions of the exponent and base.

#figure(figph[This figure shows the expression y equals log sub a of x, where y is the exponent and a is the base. Next to this expression we have x equals a to the y, where again y is the exponent and a is the base.], alt: "This figure shows the expression y equals log sub a of x, where y is the exponent and a is the base. Next to this expression we have x equals a to the y, where again y is the exponent and a is the base.", caption: none)

If we realize the logarithm is the exponent it makes the conversion easier. You may want to repeat, “base to the exponent give us the number.”

#examplebox("Example 1")[][
Convert to logarithmic form: ⓐ #math.equation(block: false, alt: "2 cubed equals 8 ,")[$2^(3) = 8 ,$] ⓑ #math.equation(block: false, alt: "5 to the power the fraction 1 over 2 equals the square root of 5 ,")[$5^(frac(1, 2)) = sqrt(5) ,$] and ⓒ #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 close parenthesis to the power 4 equals the fraction 1 over 16 .")[$attach(( frac(1, 2) ), t: 4) = frac(1, 16) .$]

#solutionbox[
#figure(figph[In part (a) we have 2 to the 3 power equals 8, where the 2 is red and the 3 is blue. Following this, we have blue y equals log sub red a of x. Then 3 equals log sub 2 of 8. Hence, if 2 cubed equals 8, then 3 equals log sub 2 of 8. In part (b) we have 5 to the 1 over 2 power equals square root of 5, where the 5 is red and the 1 over 2 is blue. Following this, we have blue y equals log sub red a of x. Then 1 over 2 equals log sub 5 of the square root of 5. Hence, if 5 to the 1 over 2 power equals the square root of 5, then 1 over 2 equals log sub 5 of the square root of 5. In part (c) we have 1 over 2 to the x power equals 1 over 16, where the 1 over 2 is red and the x is blue. Following this, we have blue y equals log sub red a of x. Then x equals log sub 1 over 2 of 1 over 16. Hence, if 1 over 2 to the x power equals 1 over 16, then x equals log sub 1 over 2 of 1 over 16.], alt: "In part (a) we have 2 to the 3 power equals 8, where the 2 is red and the 3 is blue. Following this, we have blue y equals log sub red a of x. Then 3 equals log sub 2 of 8. Hence, if 2 cubed equals 8, then 3 equals log sub 2 of 8. In part (b) we have 5 to the 1 over 2 power equals square root of 5, where the 5 is red and the 1 over 2 is blue. Following this, we have blue y equals log sub red a of x. Then 1 over 2 equals log sub 5 of the square root of 5. Hence, if 5 to the 1 over 2 power equals the square root of 5, then 1 over 2 equals log sub 5 of the square root of 5. In part (c) we have 1 over 2 to the x power equals 1 over 16, where the 1 over 2 is red and the x is blue. Following this, we have blue y equals log sub red a of x. Then x equals log sub 1 over 2 of 1 over 16. Hence, if 1 over 2 to the x power equals 1 over 16, then x equals log sub 1 over 2 of 1 over 16.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Convert to logarithmic form: ⓐ #math.equation(block: false, alt: "3 squared equals 9")[$3^(2) = 9$] ⓑ #math.equation(block: false, alt: "7 to the power the fraction 1 over 2 equals the square root of 7")[$7^(frac(1, 2)) = sqrt(7)$] ⓒ #math.equation(block: false, alt: "open parenthesis the fraction 1 over 3 close parenthesis to the power x equals the fraction 1 over 27")[$attach(( frac(1, 3) ), t: x) = frac(1, 27)$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "log sub 3 9 equals 2")[$"log"_(3) 9 = 2$] #linebreak() ⓑ #math.equation(block: false, alt: "log sub 7 the square root of 7 equals the fraction 1 over 2")[$"log"_(7) sqrt(7) = frac(1, 2)$] ⓒ #math.equation(block: false, alt: "log sub the fraction 1 over 3 the fraction 1 over 27 equals x")[$"log"_(frac(1, 3)) frac(1, 27) = x$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Convert to logarithmic form: ⓐ #math.equation(block: false, alt: "4 cubed equals 64")[$4^(3) = 64$] ⓑ #math.equation(block: false, alt: "4 to the power the fraction 1 over 3 equals the cube root of 4")[$4^(frac(1, 3)) = root(3, 4)$] ⓒ #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 close parenthesis to the power x equals the fraction 1 over 32")[$attach(( frac(1, 2) ), t: x) = frac(1, 32)$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "log sub 4 64 equals 3")[$"log"_(4) 64 = 3$] #linebreak() ⓑ #math.equation(block: false, alt: "log sub 4 the cube root of 4 equals the fraction 1 over 3")[$"log"_(4) root(3, 4) = frac(1, 3)$] ⓒ #math.equation(block: false, alt: "log sub the fraction 1 over 2 the fraction 1 over 32 equals x")[$"log"_(frac(1, 2)) frac(1, 32) = x$]
]
]

In the next example we do the reverse—convert logarithmic form to exponential form.

#examplebox("Example 2")[][
Convert to exponential form: ⓐ #math.equation(block: false, alt: "2 equals log sub 8 64 ,")[$2 = "log"_(8) 64 ,$] ⓑ #math.equation(block: false, alt: "0 equals log sub 4 1 ,")[$0 = "log"_(4) 1 ,$] and ⓒ #math.equation(block: false, alt: "minus 3 equals log sub 10 the fraction 1 over 1000 .")[$− 3 = "log"_(10) frac(1, 1000) .$]

#solutionbox[
#figure(figph[In part (a) we have 2 equals log sub 8 of 64, where the 2 is blue and the 8 is red. Following this, we have x equals red a to the blue y power. Then 64 equals 8 squared. Hence, if 2 equals log sub 8 of 64, then 64 equals 8 squared. In part (b) we have 0 equals log sub 4 of 1, where the 0 is blue and the 4 is red. Following this, we have x equals red a to the blue y power. Then 1 equals 4 to the zero power. Hence, if 0 equals log sub 4 of 1, then 1 equals 4 to the zero power. In part (c) we have negative 3 equals log sub 10 of 1 over 1000, where the negative 3 is blue and the 10 is red. Following this, we have x equals red a to the blue y power. Then 1 over 1000 equals 10 to the negative three power. Hence, if negative 3 equals log sub 10 of 1 over 1000, then 1 over 1000 equals 10 to the negative 3 power.], alt: "In part (a) we have 2 equals log sub 8 of 64, where the 2 is blue and the 8 is red. Following this, we have x equals red a to the blue y power. Then 64 equals 8 squared. Hence, if 2 equals log sub 8 of 64, then 64 equals 8 squared. In part (b) we have 0 equals log sub 4 of 1, where the 0 is blue and the 4 is red. Following this, we have x equals red a to the blue y power. Then 1 equals 4 to the zero power. Hence, if 0 equals log sub 4 of 1, then 1 equals 4 to the zero power. In part (c) we have negative 3 equals log sub 10 of 1 over 1000, where the negative 3 is blue and the 10 is red. Following this, we have x equals red a to the blue y power. Then 1 over 1000 equals 10 to the negative three power. Hence, if negative 3 equals log sub 10 of 1 over 1000, then 1 over 1000 equals 10 to the negative 3 power.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Convert to exponential form: ⓐ #math.equation(block: false, alt: "3 equals log sub 4 64")[$3 = "log"_(4) 64$] ⓑ #math.equation(block: false, alt: "0 equals log sub x 1")[$0 = "log"_(x) 1$] ⓒ #math.equation(block: false, alt: "−2 equals log sub 10 the fraction 1 over 100")[$−2 = "log"_(10) frac(1, 100)$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "64 equals 4 cubed")[$64 = 4^(3)$] #linebreak() ⓑ #math.equation(block: false, alt: "1 equals x to the power 0")[$1 = x^(0)$] ⓒ #math.equation(block: false, alt: "the fraction 1 over 100 equals 10 to the power −2")[$frac(1, 100) = 10^(−2)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Convert to exponential form: ⓐ #math.equation(block: false, alt: "3 equals log sub 3 27")[$3 = "log"_(3) 27$] ⓑ #math.equation(block: false, alt: "0 equals log sub 3 1")[$0 = "log"_(3) 1$] ⓒ #math.equation(block: false, alt: "−1 equals log sub 10 the fraction 1 over 10")[$−1 = "log"_(10) frac(1, 10)$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "27 equals 3 cubed")[$27 = 3^(3)$] ⓑ #math.equation(block: false, alt: "1 equals 3 to the power 0")[$1 = 3^(0)$] #linebreak() ⓒ #math.equation(block: false, alt: "the fraction 1 over 10 equals 10 to the power −1")[$frac(1, 10) = 10^(−1)$]
]
]


=== Evaluate Logarithmic Functions

We can solve and evaluate logarithmic equations by using the technique of converting the equation to its equivalent exponential equation.

#examplebox("Example 3")[][
Find the value of #emph[x]: ⓐ #math.equation(block: false, alt: "log sub x 36 equals 2 ,")[$"log"_(x) 36 = 2 ,$] ⓑ #math.equation(block: false, alt: "log sub 4 x equals 3 ,")[$"log"_(4) x = 3 ,$] and ⓒ #math.equation(block: false, alt: "log sub the fraction 1 over 2 the fraction 1 over 8 equals x .")[$"log"_(frac(1, 2)) frac(1, 8) = x .$]

#solutionbox[
ⓐ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "log sub x 36 equals 2")[$#h(3.8em) "log"_(x) 36 #h(0.2em) = #h(0.2em) 2$]]),
  [Convert to exponential form.], [#math.equation(block: false, alt: "x squared equals 36")[$#h(5.6em) x^(2) #h(0.2em) = #h(0.2em) 36$]],
  [Solve the quadratic.], [#math.equation(block: false, alt: "x equals 6 , x equals −6")[$#h(3.8em) x = 6 , #h(2em) cancel(x = −6)$]],
  [The base of a logarithmic function must be positive, so we eliminate #math.equation(block: false, alt: "x equals −6")[$x = −6$].], [#math.equation(block: false, alt: "x equals 6 Therefore, log sub 6 36 equals 2 .")[$#h(6.2em) x #h(0.2em) = #h(0.2em) 6 #h(2.5em) "Therefore," #h(0.2em) "log"_(6) 36 = 2 .$]],
))

ⓑ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "log sub 4 x equals 3")[$#h(6.8em) "log"_(4) x #h(0.2em) = #h(0.2em) 3$]]),
  [Convert to exponential form.], [#math.equation(block: false, alt: "4 cubed equals x")[$#h(8.2em) 4^(3) #h(0.2em) = #h(0.2em) x$]],
  [Simplify.], [#math.equation(block: false, alt: "x equals 64 Therefore, log sub 4 64 equals 3 .")[$#h(8.6em) x #h(0.2em) = #h(0.2em) 64 #h(1.3em) "Therefore," #h(0.2em) "log"_(4) 64 #h(0.2em) = #h(0.2em) 3 .$]],
))

ⓒ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "log sub the fraction 1 over 2 the fraction 1 over 8 equals x")[$"log"_(frac(1, 2)) frac(1, 8) #h(0.2em) = #h(0.2em) x$]]),
  [Convert to exponential form.], [#math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 close parenthesis to the power x equals the fraction 1 over 8")[$#h(0.4em) attach(( frac(1, 2) ), t: x) #h(0.2em) = #h(0.2em) frac(1, 8)$]],
  [Rewrite #math.equation(block: false, alt: "the fraction 1 over 8")[$frac(1, 8)$] as #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 close parenthesis cubed")[$attach(( frac(1, 2) ), t: 3)$].], [#math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 close parenthesis to the power x equals open parenthesis the fraction 1 over 2 close parenthesis cubed")[$#h(0.4em) attach(( frac(1, 2) ), t: x) #h(0.2em) = #h(0.2em) attach(( frac(1, 2) ), t: 3)$]],
  [With the same base, the exponents must be equal.], [#math.equation(block: false, alt: "x equals 3 Therefore, log sub the fraction 1 over 2 the fraction 1 over 8 equals 3")[$#h(2.4em) x #h(0.2em) = #h(0.2em) 3 #h(2em) "Therefore," #h(0.2em) "log"_(frac(1, 2)) frac(1, 8) = 3$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the value of #math.equation(block: false, alt: "x :")[$x :$] ⓐ #math.equation(block: false, alt: "log sub x 64 equals 2")[$"log"_(x) 64 = 2$] ⓑ #math.equation(block: false, alt: "log sub 5 x equals 3")[$"log"_(5) x = 3$] ⓒ #math.equation(block: false, alt: "log sub the fraction 1 over 2 the fraction 1 over 4 equals x")[$"log"_(frac(1, 2)) frac(1, 4) = x$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "x equals 8")[$x = 8$] ⓑ #math.equation(block: false, alt: "x equals 125")[$x = 125$] ⓒ #math.equation(block: false, alt: "x equals 2")[$x = 2$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the value of #math.equation(block: false, alt: "x :")[$x :$] ⓐ #math.equation(block: false, alt: "log sub x 81 equals 2")[$"log"_(x) 81 = 2$] ⓑ #math.equation(block: false, alt: "log sub 3 x equals 5")[$"log"_(3) x = 5$] ⓒ #math.equation(block: false, alt: "log sub the fraction 1 over 3 the fraction 1 over 27 equals x")[$"log"_(frac(1, 3)) frac(1, 27) = x$]

#solutionbox[
ⓐ  #linebreak()  #math.equation(block: false, alt: "x equals 9")[$x = 9$] ⓑ #math.equation(block: false, alt: "x equals 243")[$x = 243$] ⓒ #math.equation(block: false, alt: "x equals 3")[$x = 3$]
]
]

When see an expression such as #math.equation(block: false, alt: "log sub 3 27 ,")[$"log"_(3) 27 ,$] we can find its exact value two ways. By inspection we realize it means #math.equation(block: false, alt: "“ 3")[$“ 3$] to what power will be #math.equation(block: false, alt: "27 ” ?")[$27 ” ?$] Since #math.equation(block: false, alt: "3 cubed equals 27 ,")[$3^(3) = 27 ,$] we know #math.equation(block: false, alt: "log sub 3 27 equals 3 .")[$"log"_(3) 27 = 3 .$] An alternate way is to set the expression equal to #math.equation(block: false, alt: "x")[$x$] and then convert it into an exponential equation.

#examplebox("Example 4")[][
Find the exact value of each logarithm without using a calculator: ⓐ #math.equation(block: false, alt: "log sub 5 25 ,")[$"log"_(5) 25 ,$] ⓑ #math.equation(block: false, alt: "log sub 9 3 ,")[$"log"_(9) 3 ,$] and ⓒ #math.equation(block: false, alt: "log sub 2 the fraction 1 over 16 .")[$"log"_(2) frac(1, 16) .$]

#solutionbox[
ⓐ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "log sub 5 25")[$"log"_(5) 25$]]),
  [5 to what power will be 25?], [#math.equation(block: false, alt: "log sub 5 25 equals 2")[$"log"_(5) 25 #h(0.2em) = #h(0.2em) 2$]],
  [Or], [],
  [Set the expression equal to #math.equation(block: false, alt: "x")[$x$].], [#math.equation(block: false, alt: "log sub 5 25 equals x")[$"log"_(5) 25 #h(0.2em) = #h(0.2em) x$]],
  [Change to exponential form.], [#math.equation(block: false, alt: "5 to the power x equals 25")[$#h(1.8em) 5^(x) #h(0.2em) = #h(0.2em) 25$]],
  [Rewrite 25 as #math.equation(block: false, alt: "5 squared")[$5^(2)$].], [#math.equation(block: false, alt: "5 to the power x equals 5 squared")[$#h(1.8em) 5^(x) #h(0.2em) = #h(0.2em) 5^(2)$]],
  [With the same base the exponents must be equal.], [#math.equation(block: false, alt: "x equals 2 Therefore, log sub 5 25 equals 2 .")[$#h(2.3em) x #h(0.2em) = #h(0.2em) 2 #h(1em) "Therefore," #h(0.2em) "log"_(5) 25 = 2 .$]],
))

ⓑ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "log sub 9 3")[$"log"_(9) 3$]]),
  [Set the expression equal to #math.equation(block: false, alt: "x")[$x$].], [#math.equation(block: false, alt: "log sub 9 3 equals x")[$"log"_(9) 3 #h(0.7em) = #h(0.2em) x$]],
  [Change to exponential form.], [#math.equation(block: false, alt: "9 to the power x equals 3")[$#h(1.7em) 9^(x) #h(0.2em) = #h(0.2em) 3$]],
  [Rewrite 9 as #math.equation(block: false, alt: "3 squared")[$3^(2)$].], [#math.equation(block: false, alt: "open parenthesis 3 squared close parenthesis to the power x equals 3 to the power 1")[$attach(( 3^(2) ), t: x) #h(0.2em) = #h(0.2em) 3^(1)$]],
  [Simplify the exponents.], [#math.equation(block: false, alt: "3 to the power 2 x equals 3 to the power 1")[$#h(1.4em) 3^(2 x) #h(0.2em) = #h(0.2em) 3^(1)$]],
  [With the same base the exponents must be equal.], [#math.equation(block: false, alt: "2 x equals 1")[$#h(1.7em) 2 x #h(0.2em) = #h(0.2em) 1$]],
  [Solve the equation.], [#math.equation(block: false, alt: "x equals the fraction 1 over 2 Therefore, log sub 9 3 equals the fraction 1 over 2 .")[$#h(2.2em) x #h(0.2em) = #h(0.2em) frac(1, 2) #h(1em) "Therefore," #h(0.2em) "log"_(9) 3 = frac(1, 2) .$]],
))

ⓒ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "log sub 2 the fraction 1 over 16")[$"log"_(2) frac(1, 16)$]]),
  [Set the expression equal to #math.equation(block: false, alt: "x")[$x$].], [#math.equation(block: false, alt: "log sub 2 the fraction 1 over 16 equals x")[$"log"_(2) frac(1, 16) #h(0.2em) = #h(0.2em) x$]],
  [Change to exponential form.], [#math.equation(block: false, alt: "2 to the power x equals the fraction 1 over 16")[$#h(1.8em) 2^(x) #h(0.2em) = #h(0.2em) frac(1, 16)$]],
  [Rewrite 16 as #math.equation(block: false, alt: "2 to the power 4")[$2^(4)$].], [#math.equation(block: false, alt: "2 to the power x equals the fraction 1 over 2 to the power 4")[$#h(1.8em) 2^(x) #h(0.2em) = #h(0.2em) frac(1, 2^(4))$]],
  [], [#math.equation(block: false, alt: "2 to the power x equals 2 to the power −4")[$#h(1.8em) 2^(x) #h(0.2em) = #h(0.2em) 2^(−4)$]],
  [With the same base the exponents must be equal.], [#math.equation(block: false, alt: "x equals −4 Therefore, log sub 2 the fraction 1 over 16 equals −4 .")[$#h(2.3em) x #h(0.2em) = #h(0.2em) −4 #h(1em) "Therefore," #h(0.2em) "log"_(2) frac(1, 16) = −4 .$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the exact value of each logarithm without using a calculator:
ⓐ #math.equation(block: false, alt: "log sub 12 144")[$"log"_(12) 144$] ⓑ #math.equation(block: false, alt: "log sub 4 2")[$"log"_(4) 2$] ⓒ #math.equation(block: false, alt: "log sub 2 the fraction 1 over 32")[$"log"_(2) frac(1, 32)$]

#solutionbox[
ⓐ  #linebreak()  2 ⓑ #math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$] ⓒ #math.equation(block: false, alt: "−5")[$−5$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the exact value of each logarithm without using a calculator:
ⓐ #math.equation(block: false, alt: "log sub 9 81")[$"log"_(9) 81$] ⓑ #math.equation(block: false, alt: "log sub 8 2")[$"log"_(8) 2$] ⓒ #math.equation(block: false, alt: "log sub 3 the fraction 1 over 9")[$"log"_(3) frac(1, 9)$]

#solutionbox[
ⓐ 2 ⓑ #math.equation(block: false, alt: "the fraction 1 over 3")[$frac(1, 3)$] ⓒ #math.equation(block: false, alt: "−2")[$−2$]
]
]


=== Graph Logarithmic Functions

To graph a logarithmic function #math.equation(block: false, alt: "y equals log sub a x ,")[$y = "log"_(a) x ,$] it is easiest to convert the equation to its exponential form, #math.equation(block: false, alt: "x equals a to the power y .")[$x = a^(y) .$] Generally, when we look for ordered pairs for the graph of a function, we usually choose an #emph[x]-value and then determine its corresponding #emph[y]-value. In this case you may find it easier to choose #emph[y]-values and then determine its corresponding #emph[x]-value.

#examplebox("Example 5")[][
Graph #math.equation(block: false, alt: "y equals log sub 2 x .")[$y = "log"_(2) x .$]

#solutionbox[
To graph the function, we will first rewrite the logarithmic equation, #math.equation(block: false, alt: "y equals log sub 2 x ,")[$y = "log"_(2) x ,$] in exponential form, #math.equation(block: false, alt: "2 to the power y equals x .")[$2^(y) = x .$]

We will use point plotting to graph the function. It will be easier to start with values of #emph[y] and then get #emph[x].

#figure(table(
  columns: 3, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "2 to the power y equals x")[$2^(y) = x$]], [#math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$]]),
  [#math.equation(block: false, alt: "−2")[$−2$]], [#math.equation(block: false, alt: "2 to the power −2 equals the fraction 1 over 2 squared equals the fraction 1 over 4")[$2^(−2) = frac(1, 2^(2)) = frac(1, 4)$]], [#math.equation(block: false, alt: "open parenthesis the fraction 1 over 4 , minus 2 close parenthesis")[$( frac(1, 4) , − 2 )$]],
  [#math.equation(block: false, alt: "−1")[$−1$]], [#math.equation(block: false, alt: "2 to the power −1 equals the fraction 1 over 2 to the power 1 equals the fraction 1 over 2")[$2^(−1) = frac(1, 2^(1)) = frac(1, 2)$]], [#math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 , −1 close parenthesis")[$( frac(1, 2) , −1 )$]],
  [0], [#math.equation(block: false, alt: "2 to the power 0 equals 1")[$2^(0) = 1$]], [#math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$]],
  [1], [#math.equation(block: false, alt: "2 to the power 1 equals 2")[$2^(1) = 2$]], [#math.equation(block: false, alt: "open parenthesis 2 , 1 close parenthesis")[$( 2 , 1 )$]],
  [2], [#math.equation(block: false, alt: "2 squared equals 4")[$2^(2) = 4$]], [#math.equation(block: false, alt: "open parenthesis 4 , 2 close parenthesis")[$( 4 , 2 )$]],
  [3], [#math.equation(block: false, alt: "2 cubed equals 8")[$2^(3) = 8$]], [#math.equation(block: false, alt: "open parenthesis 8 , 3 close parenthesis")[$( 8 , 3 )$]],
))

#figure(figph[This figure shows the logarithmic curve going through the points (1 over 2, negative 1), (1, 0), and (2, 1).], alt: "This figure shows the logarithmic curve going through the points (1 over 2, negative 1), (1, 0), and (2, 1).", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph: #math.equation(block: false, alt: "y equals log sub 3 x .")[$y = "log"_(3) x .$]

#solutionbox[
#figure(figph[This figure shows the logarithmic curve going through the points (1 over 3, negative 1), (1, 0), and (3, 1).], alt: "This figure shows the logarithmic curve going through the points (1 over 3, negative 1), (1, 0), and (3, 1).", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph: #math.equation(block: false, alt: "y equals log sub 5 x .")[$y = "log"_(5) x .$]

#solutionbox[
#figure(figph[This figure shows the logarithmic curve going through the points (1 over 5, negative 1), (1, 0), and (5, 1).], alt: "This figure shows the logarithmic curve going through the points (1 over 5, negative 1), (1, 0), and (5, 1).", caption: none)
]
]

The graphs of #math.equation(block: false, alt: "y equals log sub 2 x ,")[$y = "log"_(2) x ,$]#math.equation(block: false, alt: "y equals log sub 3 x ,")[$y = "log"_(3) x ,$] and #math.equation(block: false, alt: "y equals log sub 5 x")[$y = "log"_(5) x$] are the shape we expect from a logarithmic function where #math.equation(block: false, alt: "a greater than 1 .")[$a > 1 .$]

We notice that for each function the graph contains the point #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis .")[$( 1 , 0 ) .$] This make sense because #math.equation(block: false, alt: "0 equals log sub a 1")[$0 = "log"_(a) 1$] means #math.equation(block: false, alt: "a to the power 0 equals 1")[$a^(0) = 1$] which is true for any #emph[a].

The graph of each function, also contains the point #math.equation(block: false, alt: "open parenthesis a , 1 close parenthesis .")[$( a , 1 ) .$] This makes sense as #math.equation(block: false, alt: "1 equals log sub a a")[$1 = "log"_(a) a$] means #math.equation(block: false, alt: "a to the power 1 equals a .")[$a^(1) = a .$] which is true for any #emph[a].

Notice too, the graph of each function #math.equation(block: false, alt: "y equals log sub a x")[$y = "log"_(a) x$] also contains the point #math.equation(block: false, alt: "open parenthesis the fraction 1 over a , −1 close parenthesis .")[$( frac(1, a) , −1 ) .$] This makes sense as #math.equation(block: false, alt: "−1 equals log sub a the fraction 1 over a")[$−1 = "log"_(a) frac(1, a)$] means #math.equation(block: false, alt: "a to the power −1 equals the fraction 1 over a ,")[$a^(−1) = frac(1, a) ,$] which is true for any #emph[a].

Look at each graph again. Now we will see that many characteristics of the logarithm function are simply ’mirror images’ of the characteristics of the corresponding exponential function.

What is the domain of the function? The graph never hits the #emph[y]-axis. The domain is all positive numbers. We write the domain in interval notation as #math.equation(block: false, alt: "open parenthesis 0 , ∞ close parenthesis .")[$( 0 , ∞ ) .$]

What is the range for each function? From the graphs we can see that the range is the set of all real numbers. There is no restriction on the range. We write the range in interval notation as #math.equation(block: false, alt: "open parenthesis − ∞ , ∞ close parenthesis .")[$( "−" ∞ , ∞ ) .$]

When the graph approaches the #emph[y]-axis so very closely but will never cross it, we call the line #math.equation(block: false, alt: "x equals 0 ,")[$x = 0 ,$] the #emph[y]-axis, a vertical asymptote.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Properties of the Graph of #math.equation(block: false, alt: "y equals log sub a x")[$y = "log"_(a) x$] when #math.equation(block: false, alt: "a greater than 1")[$a > 1$]]

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Domain], [#math.equation(block: false, alt: "open parenthesis 0 , ∞ close parenthesis")[$( 0 , ∞ )$]]),
  [Range], [#math.equation(block: false, alt: "open parenthesis − ∞ , ∞ close parenthesis")[$( "−" ∞ , ∞ )$]],
  [#math.equation(block: false, alt: "x - intercept")[$x "-" "intercept"$]], [#math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$]],
  [#math.equation(block: false, alt: "y - intercept")[$y "-" "intercept"$]], [None],
  [Contains], [#math.equation(block: false, alt: "open parenthesis a , 1 close parenthesis ,")[$( a , 1 ) ,$]#math.equation(block: false, alt: "open parenthesis the fraction 1 over a , −1 close parenthesis")[$( frac(1, a) , −1 )$]],
  [Asymptote], [#math.equation(block: false, alt: "y - axis")[$y "-" "axis"$]],
))

#figure(figph[This figure shows the logarithmic curve going through the points (1 over a, negative 1), (1, 0), and (a, 1).], alt: "This figure shows the logarithmic curve going through the points (1 over a, negative 1), (1, 0), and (a, 1).", caption: none)
]

Our next example looks at the graph of #math.equation(block: false, alt: "y equals log sub a x")[$y = "log"_(a) x$] when #math.equation(block: false, alt: "0 less than a less than 1 .")[$0 < a < 1 .$]

#examplebox("Example 6")[][
Graph #math.equation(block: false, alt: "y equals log sub the fraction 1 over 3 x .")[$y = "log"_(frac(1, 3)) x .$]

#solutionbox[
To graph the function, we will first rewrite the logarithmic equation, #math.equation(block: false, alt: "y equals log sub the fraction 1 over 3 x ,")[$y = "log"_(frac(1, 3)) x ,$] in exponential form, #math.equation(block: false, alt: "open parenthesis the fraction 1 over 3 close parenthesis to the power y equals x .")[$attach(( frac(1, 3) ), t: y) = x .$]

We will use point plotting to graph the function. It will be easier to start with values of #emph[y] and then get #emph[x].

#figure(table(
  columns: 3, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "open parenthesis the fraction 1 over 3 close parenthesis to the power y equals x")[$attach(( frac(1, 3) ), t: y) = x$]], [#math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$]]),
  [#math.equation(block: false, alt: "−2")[$−2$]], [#math.equation(block: false, alt: "open parenthesis the fraction 1 over 3 close parenthesis to the power −2 equals 3 squared equals 9")[$attach(( frac(1, 3) ), t: −2) = 3^(2) = 9$]], [#math.equation(block: false, alt: "open parenthesis 9 , −2 close parenthesis")[$( 9 , −2 )$]],
  [#math.equation(block: false, alt: "−1")[$−1$]], [#math.equation(block: false, alt: "open parenthesis the fraction 1 over 3 close parenthesis to the power −1 equals 3 to the power 1 equals 3")[$attach(( frac(1, 3) ), t: −1) = 3^(1) = 3$]], [#math.equation(block: false, alt: "open parenthesis 3 , −1 close parenthesis")[$( 3 , −1 )$]],
  [0], [#math.equation(block: false, alt: "open parenthesis the fraction 1 over 3 close parenthesis to the power 0 equals 1")[$attach(( frac(1, 3) ), t: 0) = 1$]], [#math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$]],
  [1], [#math.equation(block: false, alt: "open parenthesis the fraction 1 over 3 close parenthesis to the power 1 equals the fraction 1 over 3")[$attach(( frac(1, 3) ), t: 1) = frac(1, 3)$]], [#math.equation(block: false, alt: "open parenthesis the fraction 1 over 3 , 1 close parenthesis")[$( frac(1, 3) , 1 )$]],
  [2], [#math.equation(block: false, alt: "open parenthesis the fraction 1 over 3 close parenthesis squared equals the fraction 1 over 9")[$attach(( frac(1, 3) ), t: 2) = frac(1, 9)$]], [#math.equation(block: false, alt: "open parenthesis the fraction 1 over 9 , 2 close parenthesis")[$( frac(1, 9) , 2 )$]],
  [3], [#math.equation(block: false, alt: "open parenthesis the fraction 1 over 3 close parenthesis cubed equals the fraction 1 over 27")[$attach(( frac(1, 3) ), t: 3) = frac(1, 27)$]], [#math.equation(block: false, alt: "open parenthesis the fraction 1 over 27 , 3 close parenthesis")[$( frac(1, 27) , 3 )$]],
))

#figure(figph[This figure shows the logarithmic curve going through the points (1 over 3, 1), (1, 0), and (3, negative 1).], alt: "This figure shows the logarithmic curve going through the points (1 over 3, 1), (1, 0), and (3, negative 1).", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph: #math.equation(block: false, alt: "y equals log sub the fraction 1 over 2 x .")[$y = "log"_(frac(1, 2)) x .$]

#solutionbox[
#figure(figph[This figure shows the logarithmic curve going through the points (1 over 2, 1), (1, 0), and (2, negative 1).], alt: "This figure shows the logarithmic curve going through the points (1 over 2, 1), (1, 0), and (2, negative 1).", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph: #math.equation(block: false, alt: "y equals log sub the fraction 1 over 4 x .")[$y = "log"_(frac(1, 4)) x .$]

#solutionbox[
#figure(figph[This figure shows the logarithmic curve going through the points (1 over 4, 1), (1, 0), and (4, negative 1).], alt: "This figure shows the logarithmic curve going through the points (1 over 4, 1), (1, 0), and (4, negative 1).", caption: none)
]
]

Now, let’s look at the graphs #math.equation(block: false, alt: "y equals log sub the fraction 1 over 2 x , y equals log sub the fraction 1 over 3 x")[$y = "log"_(frac(1, 2)) x , #h(0.2em) y = "log"_(frac(1, 3)) x$] and #math.equation(block: false, alt: "y equals log sub the fraction 1 over 4 x")[$y = "log"_(frac(1, 4)) x$], so we can identify some of the properties of logarithmic functions where #math.equation(block: false, alt: "0 less than a less than 1 .")[$0 < a < 1 .$]

The graphs of all have the same basic shape. While this is the shape we expect from a logarithmic function where #math.equation(block: false, alt: "0 less than a less than 1 .")[$0 < a < 1 .$]

We notice, that for each function again, the graph contains the points,#math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis ,")[$( 1 , 0 ) ,$]#math.equation(block: false, alt: "open parenthesis a , 1 close parenthesis ,")[$( a , 1 ) ,$]#math.equation(block: false, alt: "open parenthesis the fraction 1 over a , −1 close parenthesis .")[$( frac(1, a) , −1 ) .$] This make sense for the same reasons we argued above.

We notice the domain and range are also the same—the domain is #math.equation(block: false, alt: "open parenthesis 0 , ∞ close parenthesis")[$( 0 , ∞ )$] and the range is #math.equation(block: false, alt: "open parenthesis − ∞ , ∞ close parenthesis .")[$( "−" ∞ , ∞ ) .$] The #math.equation(block: false, alt: "y")[$y$]-axis is again the vertical asymptote.

We will summarize these properties in the chart below. Which also include when #math.equation(block: false, alt: "a greater than 1 .")[$a > 1 .$]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Properties of the Graph of #math.equation(block: false, alt: "y equals log sub a x")[$y = "log"_(a) x$]]

#figure(table(
  columns: 4, align: left, inset: 6pt,
  [#strong[when] #math.equation(block: false, alt: "a greater than 1")[$a > 1$]], [#strong[when] #math.equation(block: false, alt: "0 less than a less than 1")[$0 < a < 1$]], [], [],
  [Domain], [#math.equation(block: false, alt: "open parenthesis 0 , ∞ close parenthesis")[$( 0 , ∞ )$]], [Domain], [#math.equation(block: false, alt: "open parenthesis 0 , ∞ close parenthesis")[$( 0 , ∞ )$]],
  [Range], [#math.equation(block: false, alt: "open parenthesis − ∞ , ∞ close parenthesis")[$( "−" ∞ , ∞ )$]], [Range], [#math.equation(block: false, alt: "open parenthesis − ∞ , ∞ close parenthesis")[$( "−" ∞ , ∞ )$]],
  [#math.equation(block: false, alt: "x")[$x$]-intercept], [#math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$]], [#math.equation(block: false, alt: "x")[$x$]-intercept], [#math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$]],
  [#math.equation(block: false, alt: "y")[$y$]-intercept], [none], [#math.equation(block: false, alt: "y")[$y$]-intercept], [None],
  [Contains], [#math.equation(block: false, alt: "open parenthesis a , 1 close parenthesis ,")[$( a , 1 ) ,$]#math.equation(block: false, alt: "open parenthesis the fraction 1 over a , −1 close parenthesis")[$( frac(1, a) , −1 )$]], [Contains], [#math.equation(block: false, alt: "open parenthesis a , 1 close parenthesis ,")[$( a , 1 ) ,$]#math.equation(block: false, alt: "open parenthesis the fraction 1 over a , −1 close parenthesis")[$( frac(1, a) , −1 )$]],
  [Asymptote], [#math.equation(block: false, alt: "y")[$y$]-axis], [Asymptote], [#math.equation(block: false, alt: "y")[$y$]-axis],
  [Basic shape], [increasing], [Basic shape], [Decreasing],
))

#figure(figph[This figure shows that, for a greater than 1, the logarithmic curve going through the points (1 over a, negative 1), (1, 0), and (a, 1). This figure shows that, for a greater than 0 and less than 1, the logarithmic curve going through the points (a, 1), (1, 0), and (1 over a, negative 1).], alt: "This figure shows that, for a greater than 1, the logarithmic curve going through the points (1 over a, negative 1), (1, 0), and (a, 1). This figure shows that, for a greater than 0 and less than 1, the logarithmic curve going through the points (a, 1), (1, 0), and (1 over a, negative 1).", caption: none)
]

We talked earlier about how the logarithmic function #math.equation(block: false, alt: "f to the power −1 open parenthesis x close parenthesis equals log sub a x")[$f^(−1) ( x ) = "log"_(a) x$] is the inverse of the exponential function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a to the power x .")[$f ( x ) = a^(x) .$] The graphs in  show both the exponential (blue) and logarithmic (red) functions on the same graph for both #math.equation(block: false, alt: "a greater than 1")[$a > 1$] and #math.equation(block: false, alt: "0 less than a less than 1 .")[$0 < a < 1 .$]

#figure(figph[This figure shows that, for a greater than 1, the logarithmic curve going through the points (1 over a, negative 1), (1, 0), and (a, 1). It also shows the exponential curve going through the points (1, 1 over a), (0, 1), and (1, a) along with the line y equals x. The logarithmic curve is a mirror image of the exponential curve across the y equals x line. This figure shows that, for a greater than 0 and less than 1, the logarithmic curve going through the points (a, 1), (1, 0), and (1 over a, negative 1). It also shows the exponential curve going through the points (negative 1, 1 over a), (0, 1), and (1, a) along with the line y equals x. The logarithmic curve is a mirror image of the exponential curve across the y equals x line.], alt: "This figure shows that, for a greater than 1, the logarithmic curve going through the points (1 over a, negative 1), (1, 0), and (a, 1). It also shows the exponential curve going through the points (1, 1 over a), (0, 1), and (1, a) along with the line y equals x. The logarithmic curve is a mirror image of the exponential curve across the y equals x line. This figure shows that, for a greater than 0 and less than 1, the logarithmic curve going through the points (a, 1), (1, 0), and (1 over a, negative 1). It also shows the exponential curve going through the points (negative 1, 1 over a), (0, 1), and (1, a) along with the line y equals x. The logarithmic curve is a mirror image of the exponential curve across the y equals x line.", caption: none)

Notice how the graphs are reflections of each other through the line #math.equation(block: false, alt: "y equals x .")[$y = x .$] We know this is true of inverse functions. Keeping a visual in your mind of these graphs will help you remember the domain and range of each function. Notice the #emph[x]-axis is the horizontal asymptote for the exponential functions and the #emph[y]-axis is the vertical asymptote for the logarithmic functions.


=== Solve Logarithmic Equations

When we talked about exponential functions, we introduced the number #emph[e]. Just as #emph[e] was a base for an exponential function, it can be used a base for logarithmic functions too. The logarithmic function with base #emph[e] is called the #strong[natural logarithmic function]. The function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals log sub e x")[$f ( x ) = "log"_(e) x$] is generally written #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals ln x")[$f ( x ) = "ln" #h(0.2em) x$] and we read it as “el en of #math.equation(block: false, alt: "x . ”")[$x . ”$]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Natural Logarithmic Function]

The function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals ln x")[$f ( x ) = "ln" #h(0.2em) x$] is the #strong[natural logarithmic function] with base #math.equation(block: false, alt: "e ,")[$e ,$] where #math.equation(block: false, alt: "x greater than 0 .")[$x > 0 .$]

#math.equation(block: true, alt: "y equals ln x is equivalent to x equals e to the power y")[$y = "ln" #h(0.2em) x #h(0.2em) "is equivalent to" #h(0.2em) x = e^(y)$]
]

When the base of the logarithm function is 10, we call it the #strong[common logarithmic function] and the base is not shown. If the base #emph[a] of a logarithm is not shown, we assume it is 10.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Common Logarithmic Function]

The function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals log x")[$f ( x ) = "log" #h(0.2em) x$] is the #strong[common logarithmic function] with base#math.equation(block: false, alt: "10")[$10$], where #math.equation(block: false, alt: "x greater than 0 .")[$x > 0 .$]

#math.equation(block: true, alt: "y equals log x is equivalent to x equals 10 to the power y")[$y = "log" #h(0.2em) x #h(0.2em) "is equivalent to" #h(0.2em) x = 10^(y)$]
]

#figure(figph[It will be important for you to use your calculator to evaluate both common and natural logarithms. Find the log and ln keys on your calculator.], alt: "It will be important for you to use your calculator to evaluate both common and natural logarithms. Find the log and ln keys on your calculator.", caption: none)

To solve logarithmic equations, one strategy is to change the equation to exponential form and then solve the exponential equation as we did before. As we solve logarithmic equations, #math.equation(block: false, alt: "y equals log sub a x")[$y = "log"_(a) x$], we need to remember that for the base #emph[a], #math.equation(block: false, alt: "a greater than 0")[$a > 0$] and #math.equation(block: false, alt: "a not equal to 1 .")[$a ≠ 1 .$] Also, the domain is #math.equation(block: false, alt: "x greater than 0 .")[$x > 0 .$] Just as with radical equations, we must check our solutions to eliminate any extraneous solutions.

#examplebox("Example 7")[][
Solve: ⓐ #math.equation(block: false, alt: "log sub a 49 equals 2")[$"log"_(a) 49 = 2$] and ⓑ #math.equation(block: false, alt: "ln x equals 3 .")[$"ln" #h(0.2em) x = 3 .$]

#solutionbox[
ⓐ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "log sub a 49 equals 2")[$log_(a) 49 #h(0.2em) = #h(0.2em) 2$]]),
  [Rewrite in exponential form.], [#math.equation(block: false, alt: "a squared equals 49")[$#h(1.8em) a^(2) #h(0.2em) = #h(0.2em) 49$]],
  [Solve the equation using the square root property.], [#math.equation(block: false, alt: "a equals plus or minus 7")[$#h(2.5em) a #h(0.2em) = #h(0.2em) ± 7$]],
  [The base cannot be negative, so we eliminate #math.equation(block: false, alt: "a equals minus 7.")[$a = − 7.$]], [#math.equation(block: false, alt: "a equals 7 , a equals minus 7")[$a = 7 , #h(0.5em) cancel(a = − 7)$]],
  [Check.], [],
  [#math.equation(block: false, alt: "a equals 7, log sub a 49, equals, 2; log sub 7 49, equals ?, 2; 7 squared, equals ?, 49; 49, equals, 49 ✓")[$a = 7 & log_(a) 49 & = & 2 \  & log_(7) 49 & limits(=)^(?) & 2 \  & 7^(2) & limits(=)^(?) & 49 \  & 49 & = & 49 ✓$]], [],
))

ⓑ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "ln x equals 3")[$#h(4.6em) "ln" #h(0.2em) x #h(0.2em) = #h(0.2em) 3$]]),
  [Rewrite in exponential form.], [#math.equation(block: false, alt: "e cubed equals x")[$#h(5.1em) e^(3) #h(0.2em) = #h(0.2em) x$]],
  [Check.], [],
  [#math.equation(block: false, alt: "x equals e cubed, ln x, equals, 3; ln e cubed, equals ?, 3; e cubed, equals, e cubed ✓")[$x = e^(3) & "ln" #h(0.2em) x & = & 3 \  & "ln" #h(0.2em) e^(3) & limits(=)^(?) & 3 \  & e^(3) & = & e^(3) ✓$]], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: ⓐ #math.equation(block: false, alt: "log sub a 121 equals 2")[$"log"_(a) 121 = 2$] ⓑ #math.equation(block: false, alt: "ln x equals 7")[$"ln" #h(0.2em) x = 7$]

#solutionbox[
ⓐ  #linebreak()  #math.equation(block: false, alt: "a equals 11")[$a = 11$] #linebreak() ⓑ #math.equation(block: false, alt: "x equals e to the power 7")[$x = e^(7)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: ⓐ #math.equation(block: false, alt: "log sub a 64 equals 3")[$"log"_(a) 64 = 3$] ⓑ #math.equation(block: false, alt: "ln x equals 9")[$"ln" #h(0.2em) x = 9$]

#solutionbox[
ⓐ  #linebreak()  #math.equation(block: false, alt: "a equals 4")[$a = 4$] #linebreak() ⓑ #math.equation(block: false, alt: "x equals e to the power 9")[$x = e^(9)$]
]
]

#examplebox("Example 8")[][
Solve: ⓐ #math.equation(block: false, alt: "log sub 2 open parenthesis 3 x minus 5 close parenthesis equals 4")[$"log"_(2) ( 3 x − 5 ) = 4$] and ⓑ #math.equation(block: false, alt: "ln e to the power 2 x equals 4 .")[$"ln" #h(0.2em) e^(2 x) = 4 .$]

#solutionbox[
ⓐ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "log sub 2 open parenthesis 3 x minus 5 close parenthesis equals 4")[$log_(2) ( 3 x − 5 ) #h(0.2em) = #h(0.2em) 4 #h(2.3em)$]]),
  [Rewrite in exponential form.], [#math.equation(block: false, alt: "2 to the power 4 equals 3 x minus 5")[$2^(4) #h(0.2em) = #h(0.2em) 3 x − 5$]],
  [Simplify.], [#math.equation(block: false, alt: "16 equals 3 x minus 5")[$16 #h(0.2em) = #h(0.2em) 3 x − 5$]],
  [Solve the equation.], [#math.equation(block: false, alt: "21 equals 3 x")[$21 #h(0.2em) = #h(0.2em) 3 x #h(1.7em)$]],
  [], [#math.equation(block: false, alt: "7 equals x")[$7 #h(0.2em) = #h(0.2em) x #h(2.2em)$]],
  [Check.], [],
  [#math.equation(block: false, alt: "x equals 7, log sub 2 open parenthesis 3 x minus 5 close parenthesis, equals, 4; log sub 2 open parenthesis 3 times 7 minus 5 close parenthesis, equals ?, 4; log sub 2 open parenthesis 16 close parenthesis, equals ?, 4; 2 to the power 4, equals ?, 16; 16, equals, 16 ✓")[$x = 7 & log_(2) ( 3 x − 5 ) & = & 4 \  & log_(2) ( 3 ⋅ 7 − 5 ) & limits(=)^(?) & 4 \  & log_(2) ( 16 ) & limits(=)^(?) & 4 \  & 2^(4) & limits(=)^(?) & 16 \  & 16 & = & 16 ✓ #h(0.6em)$]], [],
))

ⓑ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "ln e to the power 2 x equals 4")[$ln #h(0.2em) e^(2 x) #h(0.2em) = #h(0.2em) 4 #h(0.8em)$]]),
  [Rewrite in exponential form.], [#math.equation(block: false, alt: "e to the power 4 equals e to the power 2 x")[$e^(4) #h(0.2em) = #h(0.2em) e^(2 x)$]],
  [Since the bases are the same the exponents are equal.], [#math.equation(block: false, alt: "4 equals 2 x")[$4 #h(0.2em) = #h(0.2em) 2 x #h(0.3em)$]],
  [Solve the equation.], [#math.equation(block: false, alt: "2 equals x")[$2 #h(0.2em) = #h(0.2em) x #h(0.8em)$]],
  [Check.], [],
  [#math.equation(block: false, alt: "x equals 2, ln e to the power 2 x, equals, 4; ln e to the power 2 times 2, equals ?, 4; ln e to the power 4, equals ?, 4; e to the power 4, equals, e to the power 4 ✓")[$x = 2 & ln #h(0.2em) e^(2 x) & = & 4 \  & ln #h(0.2em) e^(2 · 2) & limits(=)^(?) & 4 \  & ln #h(0.2em) e^(4) & limits(=)^(?) & 4 \  & e^(4) & = & e^(4) ✓$]], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: ⓐ #math.equation(block: false, alt: "log sub 2 open parenthesis 5 x minus 1 close parenthesis equals 6")[$"log"_(2) ( 5 x − 1 ) = 6$] ⓑ #math.equation(block: false, alt: "ln e to the power 3 x equals 6")[$"ln" #h(0.2em) e^(3 x) = 6$]

#solutionbox[
ⓐ  #linebreak()  #math.equation(block: false, alt: "x equals 13")[$x = 13$] #linebreak() ⓑ #math.equation(block: false, alt: "x equals 2")[$x = 2$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: ⓐ #math.equation(block: false, alt: "log sub 3 open parenthesis 4 x plus 3 close parenthesis equals 3")[$"log"_(3) ( 4 x + 3 ) = 3$] ⓑ #math.equation(block: false, alt: "ln e to the power 4 x equals 4")[$"ln" #h(0.2em) e^(4 x) = 4$]

#solutionbox[
ⓐ  #linebreak()  #math.equation(block: false, alt: "x equals 6")[$x = 6$] #linebreak() ⓑ #math.equation(block: false, alt: "x equals 1")[$x = 1$]
]
]


=== Use Logarithmic Models in Applications

There are many applications that are modeled by logarithmic equations. We will first look at the logarithmic equation that gives the decibel (dB) level of sound. Decibels range from 0, which is barely audible to 160, which can rupture an eardrum. The #math.equation(block: false, alt: "10 to the power −12")[$10^(−12)$] in the formula represents the intensity of sound that is barely audible.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Decibel Level of Sound]

The loudness level, #emph[D], measured in decibels, of a sound of intensity, #emph[I], measured in watts per square inch is

#math.equation(block: true, alt: "D equals 10 log open parenthesis the fraction I over 10 to the power −12 close parenthesis")[$D = 10 #h(0.2em) "log" ( frac(I, 10^(−12)) )$]
]

#examplebox("Example 9")[][
Extended exposure to noise that measures 85 dB can cause permanent damage to the inner ear which will result in hearing loss. What is the decibel level of music coming through ear phones with intensity #math.equation(block: false, alt: "10 to the power −2")[$10^(−2)$] watts per square inch?

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[The image displays the formula D = 10 log(I / 10^-12), which calculates the decibel level (D) from a given sound intensity (I).], alt: "The image displays the formula D = 10 log(I / 10^-12), which calculates the decibel level (D) from a given sound intensity (I).", caption: none)]),
  [Substitute in the intensity level, #emph[I.]], [#figure(figph[An equation: D = 10 log(10^-2 / 10^-12), demonstrating a base-10 logarithm of a fraction of powers of ten, with 10^-2 highlighted in red.], alt: "An equation: D = 10 log(10^-2 / 10^-12), demonstrating a base-10 logarithm of a fraction of powers of ten, with 10^-2 highlighted in red.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation is displayed: D = 10 log(10^10).], alt: "A mathematical equation is displayed: D = 10 log(10^10).", caption: none)],
  [Since #math.equation(block: false, alt: "log 10 to the power 10 equals 10 .")[$"log" 10^(10) = 10 .$]], [#figure(figph[A mathematical equation shows D equals 10 times 10], alt: "A mathematical equation shows D equals 10 times 10", caption: none)],
  [Multiply.], [#figure(figph[The image displays a clear mathematical expression: D = 100. The text is in a bold, sans-serif font against a plain white background, occupying the top left portion of the frame.], alt: "The image displays a clear mathematical expression: D = 100. The text is in a bold, sans-serif font against a plain white background, occupying the top left portion of the frame.", caption: none)],
  [], [The decibel level of music coming through earphones is 100 dB.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
What is the decibel level of one of the new quiet dishwashers with intensity #math.equation(block: false, alt: "10 to the power −7")[$10^(−7)$] watts per square inch?

#solutionbox[
The quiet dishwashers have a decibel level of 50 dB.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
What is the decibel level heavy city traffic with intensity #math.equation(block: false, alt: "10 to the power −3")[$10^(−3)$] watts per square inch?

#solutionbox[
The decibel level of heavy traffic is 90 dB.
]
]

The magnitude #math.equation(block: false, alt: "R")[$R$] of an earthquake is measured by a logarithmic scale called the Richter scale. The model is #math.equation(block: false, alt: "R equals log I ,")[$R = "log" #h(0.2em) I ,$] where #math.equation(block: false, alt: "I")[$I$] is the intensity of the shock wave. This model provides a way to measure #strong[earthquake intensity].

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Earthquake Intensity]

The magnitude #emph[R] of an earthquake is measured by #math.equation(block: false, alt: "R equals log I ,")[$R = "log" #h(0.2em) I ,$] where #emph[I] is the intensity of its shock wave.
]

#examplebox("Example 10")[][
In 1906, San Francisco experienced an intense earthquake with a magnitude of 7.8 on the Richter scale. Over 80% of the city was destroyed by the resulting fires. In 2014, Los Angeles experienced a moderate earthquake that measured 5.1 on the Richter scale and caused \$108 million dollars of damage. Compare the intensities of the two earthquakes.

#solutionbox[
To compare the intensities, we first need to convert the magnitudes to intensities using the log formula. Then we will set up a ratio to compare the intensities.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Convert the magnitudes to intensities.], [#math.equation(block: false, alt: "R equals log I")[$#h(0.6em) R = "log" #h(0.2em) I$]]),
  [#math.equation(block: false, alt: "1906 earthquake")[$#h(2em) "1906 earthquake"$]], [#math.equation(block: false, alt: "7.8 equals log I")[$7.8 = "log" #h(0.2em) I$]],
  [#math.equation(block: false, alt: "Convert to exponential form.")[$#h(2em) "Convert to exponential form."$]], [#math.equation(block: false, alt: "I equals 10 to the power 7.8")[$#h(0.86em) I = 10^(7.8)$]],
  [#math.equation(block: false, alt: "2014 earthquake")[$#h(2em) "2014 earthquake"$]], [#math.equation(block: false, alt: "5.1 equals log I")[$5.1 = "log" #h(0.2em) I$]],
  [#math.equation(block: false, alt: "Convert to exponential form.")[$#h(2em) "Convert to exponential form."$]], [#math.equation(block: false, alt: "I equals 10 to the power 5.1")[$#h(0.92em) I = 10^(5.1)$]],
  [Form a ratio of the intensities.], [#math.equation(block: false, alt: "the fraction Intensity for 1906 over Intensity for 2014")[$frac("Intensity" #h(0.2em) "for" #h(0.2em) 1906, "Intensity" #h(0.2em) "for" #h(0.2em) 2014)$]],
  [Substitute in the values.], [#math.equation(block: false, alt: "the fraction 10 to the power 7.8 over 10 to the power 5.1")[$#h(1.5em) frac(10^(7.8), 10^(5.1))$]],
  [Divide by subtracting the exponents.], [#math.equation(block: false, alt: "10 to the power 2.7")[$#h(1.5em) 10^(2.7)$]],
  [Evaluate.], [#math.equation(block: false, alt: "501")[$#h(1.5em) 501$]],
  [], [The intensity of the 1906 earthquake was about 501 times the intensity of the 2014 earthquake.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
In 1906, San Francisco experienced an intense earthquake with a magnitude of 7.8 on the Richter scale. In 1989, the Loma Prieta earthquake also affected the San Francisco area, and measured 6.9 on the Richter scale. Compare the intensities of the two earthquakes.

#solutionbox[
The intensity of the 1906 earthquake was about 8 times the intensity of the 1989 earthquake.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
In 2014, Chile experienced an intense earthquake with a magnitude of 8.2 on the Richter scale. In 2014, Los Angeles also experienced an earthquake which measured 5.1 on the Richter scale. Compare the intensities of the two earthquakes.

#solutionbox[
The intensity of the earthquake in Chile was about 1,259 times the intensity of the earthquake in Los Angeles.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Access these online resources for additional instruction and practice with evaluating and graphing logarithmic functions.

- #link("https://openstax.org/l/37logasexponent")[Re-writing logarithmic equations in exponential form]
- #link("https://openstax.org/l/37Simplifylog")[Simplifying Logarithmic Expressions]
- #link("https://openstax.org/l/37Graphlog")[Graphing logarithmic functions]
- #link("https://openstax.org/l/37Finddecibel")[Using logarithms to calculate decibel levels]
]


=== Key Concepts

- #strong[Properties of the Graph of] #math.equation(block: false, alt: "y equals log sub a x :")[$y = "log"_(a) x :$] #linebreak() 
#figure(table(
  columns: 4, align: left, inset: 6pt,
  [#strong[when] #math.equation(block: false, alt: "a greater than 1")[$a > 1$]], [#strong[when] #math.equation(block: false, alt: "0 less than a less than 1")[$0 < a < 1$]], [], [],
  [Domain], [#math.equation(block: false, alt: "open parenthesis 0 , ∞ close parenthesis")[$( 0 , ∞ )$]], [Domain], [#math.equation(block: false, alt: "open parenthesis 0 , ∞ close parenthesis")[$( 0 , ∞ )$]],
  [Range], [#math.equation(block: false, alt: "open parenthesis − ∞ , ∞ close parenthesis")[$( "−" ∞ , ∞ )$]], [Range], [#math.equation(block: false, alt: "open parenthesis − ∞ , ∞ close parenthesis")[$( "−" ∞ , ∞ )$]],
  [#emph[x]-intercept], [#math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$]], [#emph[x]-intercept], [#math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$]],
  [#emph[y]-intercept], [none], [#emph[y]-intercept], [none],
  [Contains], [#math.equation(block: false, alt: "open parenthesis a , 1 close parenthesis ,")[$( a , 1 ) ,$]#math.equation(block: false, alt: "open parenthesis the fraction 1 over a , −1 close parenthesis")[$( frac(1, a) , −1 )$]], [Contains], [#math.equation(block: false, alt: "open parenthesis a , 1 close parenthesis ,")[$( a , 1 ) ,$]#math.equation(block: false, alt: "open parenthesis the fraction 1 over a , −1 close parenthesis")[$( frac(1, a) , −1 )$]],
  [Asymptote], [#emph[y]-axis], [Asymptote], [#emph[y]-axis],
  [Basic shape], [increasing], [Basic shape], [decreasing],
))

 #linebreak() 
#figure(figph[This figure shows that, for a greater than 1, the logarithmic curve going through the points (1 over a, negative 1), (1, 0), and (a, 1). This figure shows that, for a greater than 0 and less than 1, the logarithmic curve going through the points (a, 1), (1, 0), and (1 over a, negative 1).], alt: "This figure shows that, for a greater than 1, the logarithmic curve going through the points (1 over a, negative 1), (1, 0), and (a, 1). This figure shows that, for a greater than 0 and less than 1, the logarithmic curve going through the points (a, 1), (1, 0), and (1 over a, negative 1).", caption: none)
- #strong[Decibel Level of Sound:] The loudness level, #math.equation(block: false, alt: "D")[$D$], measured in decibels, of a sound of intensity, #math.equation(block: false, alt: "I")[$I$], measured in watts per square inch is #math.equation(block: false, alt: "D equals 10 log open parenthesis the fraction I over 10 to the power −12 close parenthesis .")[$D = 10 "log" ( frac(I, 10^(−12)) ) .$]
- #strong[Earthquake Intensity:] The magnitude #math.equation(block: false, alt: "R")[$R$] of an earthquake is measured by #math.equation(block: false, alt: "R equals log I ,")[$R = "log" #h(0.2em) I ,$] where #math.equation(block: false, alt: "I")[$I$] is the intensity of its shock wave.


==== Practice Makes Perfect

#strong[Convert Between Exponential and Logarithmic Form]

In the following exercises, convert from exponential to logarithmic form.

#math.equation(block: true, alt: "4 squared equals 16")[$4^(2) = 16$]

#math.equation(block: true, alt: "2 to the power 5 equals 32")[$2^(5) = 32$]

#solutionbox[
#math.equation(block: true, alt: "log sub 2 32 equals 5")[$"log"_(2) 32 = 5$]
]

#math.equation(block: true, alt: "3 cubed equals 27")[$3^(3) = 27$]

#math.equation(block: true, alt: "5 cubed equals 125")[$5^(3) = 125$]

#solutionbox[
#math.equation(block: true, alt: "log sub 5 125 equals 3")[$"log"_(5) 125 = 3$]
]

#math.equation(block: true, alt: "10 cubed equals 1000")[$10^(3) = 1000$]

#math.equation(block: true, alt: "10 to the power −2 equals the fraction 1 over 100")[$10^(−2) = frac(1, 100)$]

#solutionbox[
#math.equation(block: true, alt: "log the fraction 1 over 100 equals −2")[$"log" frac(1, 100) = −2$]
]

#math.equation(block: true, alt: "x to the power the fraction 1 over 2 equals the square root of 3")[$x^(frac(1, 2)) = sqrt(3)$]

#math.equation(block: true, alt: "x to the power the fraction 1 over 3 equals the cube root of 6")[$x^(frac(1, 3)) = root(3, 6)$]

#solutionbox[
#math.equation(block: true, alt: "log sub x the cube root of 6 equals the fraction 1 over 3")[$"log"_(x) root(3, 6) = frac(1, 3)$]
]

#math.equation(block: true, alt: "32 to the power x equals the 4th root of 32")[$32^(x) = root(4, 32)$]

#math.equation(block: true, alt: "17 to the power x equals the 5th root of 17")[$17^(x) = root(5, 17)$]

#solutionbox[
#math.equation(block: true, alt: "log sub 17 the 5th root of 17 equals x")[$"log"_(17) root(5, 17) = x$]
]

#math.equation(block: true, alt: "open parenthesis the fraction 1 over 4 close parenthesis squared equals the fraction 1 over 16")[$attach(( frac(1, 4) ), t: 2) = frac(1, 16)$]

#math.equation(block: true, alt: "open parenthesis the fraction 1 over 3 close parenthesis to the power 4 equals the fraction 1 over 81")[$attach(( frac(1, 3) ), t: 4) = frac(1, 81)$]

#solutionbox[
#math.equation(block: true, alt: "log sub the fraction 1 over 3 the fraction 1 over 81 equals 4")[$"log"_(frac(1, 3)) frac(1, 81) = 4$]
]

#math.equation(block: true, alt: "3 to the power −2 equals the fraction 1 over 9")[$3^(−2) = frac(1, 9)$]

#math.equation(block: true, alt: "4 to the power −3 equals the fraction 1 over 64")[$4^(−3) = frac(1, 64)$]

#solutionbox[
#math.equation(block: true, alt: "log sub 4 the fraction 1 over 64 equals −3")[$"log"_(4) frac(1, 64) = −3$]
]

#math.equation(block: true, alt: "e to the power x equals 6")[$e^(x) = 6$]

#math.equation(block: true, alt: "e cubed equals x")[$e^(3) = x$]

#solutionbox[
#math.equation(block: true, alt: "ln x equals 3")[$"ln" #h(0.2em) x = 3$]
]

In the following exercises, convert each logarithmic equation to exponential form.

#math.equation(block: true, alt: "3 equals log sub 4 64")[$3 = "log"_(4) 64$]

#math.equation(block: true, alt: "6 equals log sub 2 64")[$6 = "log"_(2) 64$]

#solutionbox[
#math.equation(block: true, alt: "64 equals 2 to the power 6")[$64 = 2^(6)$]
]

#math.equation(block: true, alt: "4 equals log sub x 81")[$4 = "log"_(x) 81$]

#math.equation(block: true, alt: "5 equals log sub x 32")[$5 = "log"_(x) 32$]

#solutionbox[
#math.equation(block: true, alt: "32 equals x to the power 5")[$32 = x^(5)$]
]

#math.equation(block: true, alt: "0 equals log sub 12 1")[$0 = "log"_(12) 1$]

#math.equation(block: true, alt: "0 equals log sub 7 1")[$0 = "log"_(7) 1$]

#solutionbox[
#math.equation(block: true, alt: "1 equals 7 to the power 0")[$1 = 7^(0)$]
]

#math.equation(block: true, alt: "1 equals log sub 3 3")[$1 = "log"_(3) 3$]

#math.equation(block: true, alt: "1 equals log sub 9 9")[$1 = "log"_(9) 9$]

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

#math.equation(block: true, alt: "−4 equals log sub 10 the fraction 1 over 10,000")[$−4 = "log"_(10) frac(1, 10","000)$]

#math.equation(block: true, alt: "3 equals log sub 10 1,000")[$3 = "log"_(10) 1,000$]

#solutionbox[
#math.equation(block: true, alt: "1,000 equals 10 cubed")[$1,000 = 10^(3)$]
]

#math.equation(block: true, alt: "5 equals log sub e x")[$5 = "log"_(e) x$]

#math.equation(block: true, alt: "x equals log sub e 43")[$x = "log"_(e) 43$]

#solutionbox[
#math.equation(block: true, alt: "43 equals e to the power x")[$43 = e^(x)$]
]

#strong[Evaluate Logarithmic Functions]

In the following exercises, find the value of #math.equation(block: false, alt: "x")[$x$] in each logarithmic equation.

#math.equation(block: true, alt: "log sub x 49 equals 2")[$"log"_(x) 49 = 2$]

#math.equation(block: true, alt: "log sub x 121 equals 2")[$"log"_(x) 121 = 2$]

#solutionbox[
#math.equation(block: true, alt: "x equals 11")[$x = 11$]
]

#math.equation(block: true, alt: "log sub x 27 equals 3")[$"log"_(x) 27 = 3$]

#math.equation(block: true, alt: "log sub x 64 equals 3")[$"log"_(x) 64 = 3$]

#solutionbox[
#math.equation(block: true, alt: "x equals 4")[$x = 4$]
]

#math.equation(block: true, alt: "log sub 3 x equals 4")[$"log"_(3) x = 4$]

#math.equation(block: true, alt: "log sub 5 x equals 3")[$"log"_(5) x = 3$]

#solutionbox[
#math.equation(block: true, alt: "x equals 125")[$x = 125$]
]

#math.equation(block: true, alt: "log sub 2 x equals −6")[$"log"_(2) x = −6$]

#math.equation(block: true, alt: "log sub 3 x equals −5")[$"log"_(3) x = −5$]

#solutionbox[
#math.equation(block: true, alt: "x equals the fraction 1 over 243")[$x = frac(1, 243)$]
]

#math.equation(block: true, alt: "log sub the fraction 1 over 4 the fraction 1 over 16 equals x")[$"log"_(frac(1, 4)) frac(1, 16) = x$]

#math.equation(block: true, alt: "log sub the fraction 1 over 3 the fraction 1 over 9 equals x")[$"log"_(frac(1, 3)) frac(1, 9) = x$]

#solutionbox[
#math.equation(block: true, alt: "x equals 2")[$x = 2$]
]

#math.equation(block: true, alt: "log sub the fraction 1 over 4 64 equals x")[$"log"_(frac(1, 4)) 64 = x$]

#math.equation(block: true, alt: "log sub the fraction 1 over 9 81 equals x")[$"log"_(frac(1, 9)) 81 = x$]

#solutionbox[
#math.equation(block: true, alt: "x equals −2")[$x = −2$]
]

In the following exercises, find the exact value of each logarithm without using a calculator.

#math.equation(block: true, alt: "log sub 7 49")[$"log"_(7) 49$]

#math.equation(block: true, alt: "log sub 6 36")[$"log"_(6) 36$]

#solutionbox[
2
]

#math.equation(block: true, alt: "log sub 4 1")[$"log"_(4) 1$]

#math.equation(block: true, alt: "log sub 5 1")[$"log"_(5) 1$]

#solutionbox[
0
]

#math.equation(block: true, alt: "log sub 16 4")[$"log"_(16) 4$]

#math.equation(block: true, alt: "log sub 27 3")[$"log"_(27) 3$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 1 over 3")[$frac(1, 3)$]
]

#math.equation(block: true, alt: "log sub the fraction 1 over 2 2")[$"log"_(frac(1, 2)) 2$]

#math.equation(block: true, alt: "log sub the fraction 1 over 2 4")[$"log"_(frac(1, 2)) 4$]

#solutionbox[
#math.equation(block: true, alt: "−2")[$−2$]
]

#math.equation(block: true, alt: "log sub 2 the fraction 1 over 16")[$"log"_(2) frac(1, 16)$]

#math.equation(block: true, alt: "log sub 3 the fraction 1 over 27")[$"log"_(3) frac(1, 27)$]

#solutionbox[
#math.equation(block: true, alt: "−3")[$−3$]
]

#math.equation(block: true, alt: "log sub 4 the fraction 1 over 16")[$"log"_(4) frac(1, 16)$]

#math.equation(block: true, alt: "log sub 9 the fraction 1 over 81")[$"log"_(9) frac(1, 81)$]

#solutionbox[
#math.equation(block: true, alt: "−2")[$−2$]
]

#strong[Graph Logarithmic Functions]

In the following exercises, graph each logarithmic function.

#math.equation(block: true, alt: "y equals log sub 2 x")[$y = "log"_(2) x$]

#math.equation(block: true, alt: "y equals log sub 4 x")[$y = "log"_(4) x$]

#solutionbox[
#figure(figph[This figure shows the logarithmic curve going through the points (1 over 4, negative 1), (1, 0), and (4, 1).], alt: "This figure shows the logarithmic curve going through the points (1 over 4, negative 1), (1, 0), and (4, 1).", caption: none)
]

#math.equation(block: true, alt: "y equals log sub 6 x")[$y = "log"_(6) x$]

#math.equation(block: true, alt: "y equals log sub 7 x")[$y = "log"_(7) x$]

#solutionbox[
#figure(figph[This figure shows that the logarithmic curve going through the points (1 over 7, negative 1), (1, 0), and (7, 1).], alt: "This figure shows that the logarithmic curve going through the points (1 over 7, negative 1), (1, 0), and (7, 1).", caption: none)
]

#math.equation(block: true, alt: "y equals log sub 1.5 x")[$y = "log"_(1.5) x$]

#math.equation(block: true, alt: "y equals log sub 2.5 x")[$y = "log"_(2.5) x$]

#solutionbox[
#figure(figph[This figure shows the logarithmic curve going through the points (2 over 5, negative 1), (1, 0), and (2.5, 1).], alt: "This figure shows the logarithmic curve going through the points (2 over 5, negative 1), (1, 0), and (2.5, 1).", caption: none)
]

#math.equation(block: true, alt: "y equals log sub the fraction 1 over 3 x")[$y = "log"_(frac(1, 3)) x$]

#math.equation(block: true, alt: "y equals log sub the fraction 1 over 5 x")[$y = "log"_(frac(1, 5)) x$]

#solutionbox[
#figure(figph[This figure shows the logarithmic curve going through the points (1 over 5, 1), (1, 0), and (5, negative 1).], alt: "This figure shows the logarithmic curve going through the points (1 over 5, 1), (1, 0), and (5, negative 1).", caption: none)
]

#math.equation(block: true, alt: "y equals log sub 0.4 x")[$y = "log"_(0.4) x$]

#math.equation(block: true, alt: "y equals log sub 0.6 x")[$y = "log"_(0.6) x$]

#solutionbox[
#figure(figph[This figure shows the logarithmic curve going through the points (3 over 5, 1), (1, 0), and (5 over 3, negative 1).], alt: "This figure shows the logarithmic curve going through the points (3 over 5, 1), (1, 0), and (5 over 3, negative 1).", caption: none)
]

#strong[Solve Logarithmic Equations]

In the following exercises, solve each logarithmic equation.

#math.equation(block: true, alt: "log sub a 16 equals 2")[$"log"_(a) 16 = 2$]

#math.equation(block: true, alt: "log sub a 81 equals 2")[$"log"_(a) 81 = 2$]

#solutionbox[
#math.equation(block: true, alt: "a equals 9")[$a = 9$]
]

#math.equation(block: true, alt: "log sub a 8 equals 3")[$"log"_(a) 8 = 3$]

#math.equation(block: true, alt: "log sub a 27 equals 3")[$"log"_(a) 27 = 3$]

#solutionbox[
#math.equation(block: true, alt: "a equals 3")[$a = 3$]
]

#math.equation(block: true, alt: "log sub a 32 equals 2")[$"log"_(a) 32 = 2$]

#math.equation(block: true, alt: "log sub a 24 equals 3")[$"log"_(a) 24 = 3$]

#solutionbox[
#math.equation(block: true, alt: "a equals 2 the cube root of 3")[$a = 2 root(3, 3)$]
]

#math.equation(block: true, alt: "ln x equals 5")[$"ln" #h(0.2em) x = 5$]

#math.equation(block: true, alt: "ln x equals 4")[$"ln" #h(0.2em) x = 4$]

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

#math.equation(block: true, alt: "log sub 2 open parenthesis 5 x plus 1 close parenthesis equals 4")[$"log"_(2) ( 5 x + 1 ) = 4$]

#math.equation(block: true, alt: "log sub 2 open parenthesis 6 x plus 2 close parenthesis equals 5")[$"log"_(2) ( 6 x + 2 ) = 5$]

#solutionbox[
#math.equation(block: true, alt: "x equals 5")[$x = 5$]
]

#math.equation(block: true, alt: "log sub 3 open parenthesis 4 x minus 3 close parenthesis equals 2")[$"log"_(3) ( 4 x − 3 ) = 2$]

#math.equation(block: true, alt: "log sub 3 open parenthesis 5 x minus 4 close parenthesis equals 4")[$"log"_(3) ( 5 x − 4 ) = 4$]

#solutionbox[
#math.equation(block: true, alt: "x equals 17")[$x = 17$]
]

#math.equation(block: true, alt: "log sub 4 open parenthesis 5 x plus 6 close parenthesis equals 3")[$"log"_(4) ( 5 x + 6 ) = 3$]

#math.equation(block: true, alt: "log sub 4 open parenthesis 3 x minus 2 close parenthesis equals 2")[$"log"_(4) ( 3 x − 2 ) = 2$]

#solutionbox[
#math.equation(block: true, alt: "x equals 6")[$x = 6$]
]

#math.equation(block: true, alt: "ln e to the power 4 x equals 8")[$"ln" #h(0.2em) e^(4 x) = 8$]

#math.equation(block: true, alt: "ln e to the power 2 x equals 6")[$"ln" #h(0.2em) e^(2 x) = 6$]

#solutionbox[
#math.equation(block: true, alt: "x equals 3")[$x = 3$]
]

#math.equation(block: true, alt: "log x squared equals 2")[$"log" x^(2) = 2$]

#math.equation(block: true, alt: "log open parenthesis x squared minus 25 close parenthesis equals 2")[$"log" ( x^(2) − 25 ) = 2$]

#solutionbox[
#math.equation(block: true, alt: "x equals −5 the square root of 5 , x equals 5 the square root of 5")[$x = −5 sqrt(5) , x = 5 sqrt(5)$]
]

#math.equation(block: true, alt: "log sub 2 open parenthesis x squared minus 4 close parenthesis equals 5")[$"log"_(2) ( x^(2) − 4 ) = 5$]

#math.equation(block: true, alt: "log sub 3 open parenthesis x squared plus 2 close parenthesis equals 3")[$"log"_(3) ( x^(2) + 2 ) = 3$]

#solutionbox[
#math.equation(block: true, alt: "x equals −5 , x equals 5")[$x = −5 , x = 5$]
]

#strong[Use Logarithmic Models in Applications]

In the following exercises, use a logarithmic model to solve.

What is the decibel level of normal conversation with intensity #math.equation(block: false, alt: "10 to the power −6")[$10^(−6)$] watts per square inch?

What is the decibel level of a whisper with intensity #math.equation(block: false, alt: "10 to the power −10")[$10^(−10)$] watts per square inch?

#solutionbox[
A whisper has a decibel level of 20 dB.
]

What is the decibel level of the noise from a motorcycle with intensity #math.equation(block: false, alt: "10 to the power −2")[$10^(−2)$] watts per square inch?

What is the decibel level of the sound of a garbage disposal with intensity #math.equation(block: false, alt: "10 to the power −2")[$10^(−2)$] watts per square inch?

#solutionbox[
The sound of a garbage disposal has a decibel level of 100 dB.
]

In 2014, Chile experienced an intense earthquake with a magnitude of #math.equation(block: false, alt: "8.2")[$8.2$] on the Richter scale. In 2010, Haiti also experienced an intense earthquake which measured #math.equation(block: false, alt: "7.0")[$7.0$] on the Richter scale. Compare the intensities of the two earthquakes.

The Los Angeles area experiences many earthquakes. In 1994, the Northridge earthquake measured magnitude of #math.equation(block: false, alt: "6.7")[$6.7$] on the Richter scale. In 2014, Los Angeles also experienced an earthquake which measured #math.equation(block: false, alt: "5.1")[$5.1$] on the Richter scale. Compare the intensities of the two earthquakes.

#solutionbox[
The intensity of the 1994 Northridge earthquake in the Los Angeles area was about 40 times the intensity of the 2014 earthquake.
]


==== Writing Exercises

Explain how to change an equation from logarithmic form to exponential form.

Explain the difference between common logarithms and natural logarithms.

#solutionbox[
Answers will vary.
]

Explain why #math.equation(block: false, alt: "log sub a a to the power x equals x .")[$"log"_(a) a^(x) = x .$]

Explain how to find the #math.equation(block: false, alt: "log sub 7 32")[$"log"_(7) 32$] on your calculator.

#solutionbox[
Answers will vary.
]


==== Self Check

ⓐ  #linebreak()  After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

#figure(figph[This table has four rows and five columns. The first row, which serves as a header, reads I can…, Confidently, With some help, and No—I don’t get it. The first column below the header row reads Convert between exponential and logarithmic form, evaluate logarithmic functions, graph logarithmic functions, solve logarithmic equations, and use logarithmic models in applications. The rest of the cells are blank.], alt: "This table has four rows and five columns. The first row, which serves as a header, reads I can…, Confidently, With some help, and No—I don’t get it. The first column below the header row reads Convert between exponential and logarithmic form, evaluate logarithmic functions, graph logarithmic functions, solve logarithmic equations, and use logarithmic models in applications. The rest of the cells are blank.", caption: none)

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?