7.4 Equations and Inequalities involving Exponential Functions

In this section we will develop techniques for solving equations involving exponential functions. Consider the equation $2^{x} = 128$ . After a moment's calculation, we find $128 = 2^{7}$ , so we have $2^{x} = 2^{7}$ . The one-to-one property of exponential functions, detailed in Theorem 7.5, tells us that $2^{x} = 2^{7}$ if and only if $x equals 7$ . This means that not only is $x equals 7$ a solution to $2^{x} = 2^{7}$ , it is the only solution.

Now suppose we change the problem ever so slightly to $2^{x} = 129$ . We could use one of the inverse properties of exponentials and logarithms listed in Theorem 7.4 to write $129 = 2^{\log_{2} (129)}$ . We'd then have $2^{x} = 2^{\log_{2} (129)}$ , which means our solution is $x equals log sub 2 ⁡ open parenthesis 129 close parenthesis$ .

After all, the definition of $log sub 2 ⁡ open parenthesis 129 close parenthesis$ is `the exponent we put on $2$ to get $129$ .' Indeed we could have obtained this solution directly by rewriting the equation $2^{x} = 129$ in its logarithmic form $log sub 2 ⁡ open parenthesis 129 close parenthesis equals x$ . Either way, in order to get a reasonable decimal approximation to this number, we'd use the change of base formula, Theorem 7.8, to give us something more calculator friendly. Typically this means we convert our answer to base 10 or base $e$ , and we choose the latter: $\log_{2} (129) = \frac{\ln (129)}{\ln (2)} \approx 7.011$ .

Still another way to obtain this answer is to `take the natural log' of both sides of the equation. Since $f (x) = \ln (x)$ is a function, as long as two quantities are equal, their natural logs are equal.¹

We then use the Power Rule to write the exponent $x$ as a factor then divide both sides by the constant $ln ⁡ open parenthesis 2 close parenthesis$ to obtain our answer.²

$\begin{matrix} 2^{x} & = & 129 \\ \ln (2^{x}) & = & \ln (129) & Take the natural log of both sides. \\ x \ln (2) & = & \ln (129) & Power Rule \\ x & = & \frac{\ln (129)}{\ln (2)} \end{matrix}$

We summarize our two strategies for solving equations featuring exponential functions below.

Steps for Solving an Equation involving Exponential Functions

Isolate the exponential function.
1. If convenient, express both sides with a common base and equate the exponents.
2. Otherwise, take the natural log of both sides of the equation and use the Power Rule.

Example 1

Solve the following equations. Check your answer using a graphing utility.

$2^{3 x} = 16^{1 - x}$
$2000 = 1000 \cdot 3^{- 0.1 t}$
$9 \cdot 3^{x} = 7^{2 x}$
$75 = \frac{100}{1 + 3 e^{- 2 t}}$
$25^{x} = 5^{x} + 6$
$\frac{e^{x} - e^{- x}}{2} = 5$

Since $16$ is a power of $2$ , we can rewrite $2^{3 x} = 16^{1 - x}$ as $2^{3 x} = {(2^{4})}^{1 - x}$ . Using properties of exponents, we get $2^{3 x} = 2^{4 (1 - x)}$ .
Using the one-to-one property of exponential functions, we get $3 x = 4 (1 - x)$ which gives $x = \frac{4}{7}$ .
Graphing $f (x) = 2^{3 x}$ and $g (x) = 16^{1 - x}$ and see that they intersect at $x \approx 0.571 \approx \frac{4}{7}$ .
We begin solving $2000 = 1000 \cdot 3^{- 0.1 t}$ by dividing both sides by $1000$ to isolate the exponential which yields $3^{- 0.1 t} = 2$ .
Since it is inconvenient to write $2$ as a power of $3$ , we use the natural log to get $\ln (3^{- 0.1 t}) = \ln (2)$ .
Using the Power Rule, we get $- 0.1 t \ln (3) = \ln (2)$ , so we divide both sides by $minus 0.1 ⁢ ln ⁡ open parenthesis 3 close parenthesis$ and obtain $t = - \frac{\ln (2)}{0.1 \ln (3)} = - \frac{10 \ln (2)}{\ln (3)}$ .
We see the graphs of $f ⁢ open parenthesis x close parenthesis equals 2000$ and $g (x) = 1000 \cdot 3^{- 0.1 x}$ intersect at $x \approx - 6.309 \approx - \frac{10 \ln (2)}{\ln (3)}$ .
Figure 7.63
Figure 7.64
Checking $2^{3 x} = 16^{1 - x}$
Checking $2000 = 1000 \cdot 3^{- 0.1 t}$
We first note that we can rewrite the equation $9 \cdot 3^{x} = 7^{2 x}$ as $3^{2} \cdot 3^{x} = 7^{2 x}$ to obtain $3^{x + 2} = 7^{2 x}$ .
Since it is not convenient to express both sides as a power of $3$ (or $7$ for that matter) we use the natural log: $\ln (3^{x + 2}) = \ln (7^{2 x})$ .
The power rule gives $(x + 2) \ln (3) = 2 x \ln (7)$ . Even though this equation appears very complicated, keep in mind that $ln ⁡ open parenthesis 3 close parenthesis$ and $ln ⁡ open parenthesis 7 close parenthesis$ are just constants.
The equation $(x + 2) \ln (3) = 2 x \ln (7)$ is actually a linear equation (do you see why?) and as such we gather all of the terms with $x$ on one side, and the constants on the other. We then divide both sides by the coefficient of $x$ , which we obtain by factoring.
$\begin{matrix} (x + 2) \ln (3) & = & 2 x \ln (7) \\ x \ln (3) + 2 \ln (3) & = & 2 x \ln (7) \\ 2 \ln (3) & = & 2 x \ln (7) - x \ln (3) \\ 2 \ln (3) & = & x (2 \ln (7) - \ln (3)) & Factor. \\ x & = & \frac{2 \ln (3)}{2 \ln (7) - \ln (3)} \end{matrix}$
We see the graphs of $f (x) = 9 \cdot 3^{x}$ and $g (x) = 7^{2 x}$ intersect at $x \approx 0.787 \approx \frac{2 \ln (3)}{2 \ln (7) - \ln (3)}$ .
Our objective in solving $75 = \frac{100}{1 + 3 e^{- 2 t}}$ is to first isolate the exponential.
To that end, we clear denominators and get $75 (1 + 3 e^{- 2 t}) = 100$ , or $75 + 225 e^{- 2 t} = 100$ . We get $225 e^{- 2 t} = 25$ , so finally, $e^{- 2 t} = \frac{1}{9}$ .
Taking the natural log of both sides gives $\ln (e^{- 2 t}) = \ln (\frac{1}{9})$ . Since natural log is log base $e$ , $\ln (e^{- 2 t}) = - 2 t$ . Likewise, we use the Power Rule to rewrite $\ln (\frac{1}{9}) = - \ln (9)$ .
Putting these two steps together, we simplify $\ln (e^{- 2 t}) = \ln (\frac{1}{9})$ to $- 2 t = - \ln (9)$ . We arrive at our solution, $t = \frac{\ln (9)}{2}$ which simplifies to $t equals ln ⁡ open parenthesis 3 close parenthesis$ . (Can you explain why?)
To check, we see the graphs of $f ⁢ open parenthesis x close parenthesis equals 75$ and $g (x) = \frac{100}{1 + 3 e^{- 2 x}}$ , intersect at $x \approx 1.099 \approx \ln (3)$ .
Figure 7.65
Figure 7.66
Checking $9 \cdot 3^{x} = 7^{2 x}$
Checking $75 = \frac{100}{1 + 3 e^{- 2 t}}$
We start solving $25^{x} = 5^{x} + 6$ by rewriting $25 equals 5 squared$ so that we have ${(5^{2})}^{x} = 5^{x} + 6$ , or $5^{2 x} = 5^{x} + 6$ .
Even though we have a common base, having two terms on the right hand side of the equation foils our plan of equating exponents or taking logs.
If we stare at this long enough, we notice that we have three terms with the exponent on one term exactly twice that of another. To our surprise and delight, we have a `quadratic in disguise'.
Letting $u = 5^{x}$ , we have $u^{2} = {(5^{x})}^{2} = 5^{2 x}$ so the equation $5^{2 x} = 5^{x} + 6$ becomes $u^{2} = u + 6$ . Solving this as $u^{2} - u - 6 = 0$ gives $u = - 2$ or $u equals 3$ . Since $u = 5^{x}$ , we have $5^{x} = - 2$ or $5^{x} = 3$ .
Since $5^{x} = - 2$ has no real solution,³ we focus on $5^{x} = 3$ . Since it isn't convenient to express $3$ as a power of $5$ , we take natural logs and get $\ln (5^{x}) = \ln (3)$ so that $x ⁢ ln ⁡ open parenthesis 5 close parenthesis equals ln ⁡ open parenthesis 3 close parenthesis$ or $x = \frac{\ln (3)}{\ln (5)}$ .
We see the graphs of $f (x) = 25^{x}$ and $g (x) = 5^{x} + 6$ intersect at $x \approx 0.683 \approx \frac{\ln (3)}{\ln (5)}$ .
Clearing the denominator in $\frac{e^{x} - e^{- x}}{2} = 5$ gives $e^{x} - e^{- x} = 10$ , at which point we pause to consider how to proceed. Rewriting $e^{- x} = \frac{1}{e^{x}}$ , we see we have another denominator to clear: $e^{x} - \frac{1}{e^{x}} = 10$ .
Doing so gives $e^{2 x} - 1 = 10 e^{x}$ , which, once again fits the criteria of being a `quadratic in disguise.'
If we let $u = e^{x}$ , then $u^{2} = e^{2 x}$ so the equation $e^{2 x} - 1 = 10 e^{x}$ can be viewed as $u^{2} - 1 = 10 u$ . Solving $u^{2} - 10 u - 1 = 0$ using the quadratic formula gives $u = 5 \pm \sqrt{26}$ .
From this, we have $e^{x} = 5 \pm \sqrt{26}$ . Since $5 - \sqrt{26} < 0$ , we get no real solution to $e^{x} = 5 - \sqrt{26}$ (why not?) but for $e^{x} = 5 + \sqrt{26}$ , we take natural logs to obtain $x = \ln (5 + \sqrt{26})$ .
We see the graphs of $f (x) = \frac{e^{x} - e^{- x}}{2}$ and $g ⁢ open parenthesis x close parenthesis equals 5$ intersect at $x \approx 2.312 \approx \ln (5 + \sqrt{26})$ .
Figure 7.67
Figure 7.68
Checking $25^{x} = 5^{x} + 6$
Checking $\frac{e^{x} - e^{- x}}{2} = 5$

Note that verifying our solutions to the equations in Example 7.4.1 analytically holds great educational value, since it reviews many of the properties of logarithms and exponents in tandem.

For example, to verify our solution to $2000 = 1000 \cdot 3^{- 0.1 t}$ , we substitute $t = - \frac{10 \ln (2)}{\ln (3)}$ and check:

$\begin{matrix} 2000 & \overset{?}{=} & 1000 \cdot 3^{- 0.1 (- \frac{10 \ln (2)}{\ln (3)})} \\ 2000 & \overset{?}{=} & 1000 \cdot 3^{\frac{\ln (2)}{\ln (3)}} \\ 2000 & \overset{?}{=} & 1000 \cdot 3^{\log_{3} (2)} & Change of Base \\ 2000 & \overset{?}{=} & 1000 \cdot 2 & Inverse Property \\ 2000 & \overset{✓}{=} & 2000 \end{matrix}$

We strongly encourage the reader to check the remaining equations analytically as well.

Since exponential functions are continuous on their domains, the Intermediate Value Theorem 2.14 applies. This allows us to solve inequalities using sign diagrams as demonstrated below.

Example 2

Solve the following inequalities. Check your answer graphically.

$2^{x^{2} - 3 x} - 16 \geq 0$
$\frac{e^{x}}{e^{x} - 4} \leq 3$
$t ⁢ e to the power 2 ⁢ t less than 4 ⁢ t$

Since we already have $0$ on one side of the inequality, we set $r (x) = 2^{x^{2} - 3 x} - 16$ .
The domain of $r$ is all real numbers, so to construct our sign diagram, we need to find the zeros of $r$ .
Setting $r ⁢ open parenthesis x close parenthesis equals 0$ gives $2^{x^{2} - 3 x} - 16 = 0$ or $2^{x^{2} - 3 x} = 16$ . Since $16 = 2^{4}$ we have $2^{x^{2} - 3 x} = 2^{4}$ . By the one-to-one property of exponential functions, $x^{2} - 3 x = 4$ which gives $x equals 4$ and $x = - 1$ .
From the sign diagram, we see $r ⁢ open parenthesis x close parenthesis greater than or equal to 0$ on $(- \infty, - 1] \cup [4, \infty)$ , which is our solution.
Graphing $r (x) = 2^{x^{2} - 3 x} - 16$ , we find it is on or above the line $y equals 0$ (the $x$ -axis) precisely on the intervals $(- \infty, - 1]$ and $open bracket 4, ∞ close parenthesis$ which checks our answer.
Figure 7.69
Figure 7.70
A Sign Diagram for $r (x) = 2^{x^{2} - 3 x} - 16$
Checking $2^{x^{2} - 3 x} - 16 \geq 0$
The first step we need to take to solve $\frac{e^{x}}{e^{x} - 4} \leq 3$ is to get $0$ on one side of the inequality. To that end, we subtract $3$ from both sides and get a common denominator
$\begin{matrix} \frac{e^{x}}{e^{x} - 4} & \leq & 3 \\ \frac{e^{x}}{e^{x} - 4} - 3 & \leq & 0 \\ \frac{e^{x}}{e^{x} - 4} - \frac{3 (e^{x} - 4)}{e^{x} - 4} & \leq & 0 & Common denomintors. \\ \frac{12 - 2 e^{x}}{e^{x} - 4} & \leq & 0 \end{matrix}$
We set $r (x) = \frac{12 - 2 e^{x}}{e^{x} - 4}$ and we note that $r$ is undefined when its denominator $e^{x} - 4 = 0$ , or when $e^{x} = 4$ . Solving this gives $x equals ln ⁡ open parenthesis 4 close parenthesis$ , so the domain of $r$ is $open parenthesis minus ∞, ln ⁡ open parenthesis 4 close parenthesis close parenthesis union open parenthesis ln ⁡ open parenthesis 4 close parenthesis, ∞ close parenthesis$ .
To find the zeros of $r$ , we solve $r ⁢ open parenthesis x close parenthesis equals 0$ and obtain $12 - 2 e^{x} = 0$ . We find $e^{x} = 6$ , or $x equals ln ⁡ open parenthesis 6 close parenthesis$ .
When we build our sign diagram, finding test values may be a little tricky since we need to check values around $ln ⁡ open parenthesis 4 close parenthesis$ and $ln ⁡ open parenthesis 6 close parenthesis$ .
Recall that the function $\ln (x)$ is increasing⁴ which means $ln ⁡ open parenthesis 3 close parenthesis less than ln ⁡ open parenthesis 4 close parenthesis less than ln ⁡ open parenthesis 5 close parenthesis less than ln ⁡ open parenthesis 6 close parenthesis less than ln ⁡ open parenthesis 7 close parenthesis$ .
To determine the sign of $r ⁢ open parenthesis ln ⁡ open parenthesis 3 close parenthesis close parenthesis$ , we remember that $e^{\ln (3)} = 3$ and get
$r (\ln (3)) = \frac{12 - 2 e^{\ln (3)}}{e^{\ln (3)} - 4} = \frac{12 - 2 (3)}{3 - 4} = - 6 .$
We determine the signs of $r ⁢ open parenthesis ln ⁡ open parenthesis 5 close parenthesis close parenthesis$ and $r ⁢ open parenthesis ln ⁡ open parenthesis 7 close parenthesis close parenthesis$ similarly.⁵ From the sign diagram, we find our answer to be $open parenthesis minus ∞, ln ⁡ open parenthesis 4 close parenthesis close parenthesis union open bracket ln ⁡ open parenthesis 6 close parenthesis, ∞ close parenthesis$ .
Using a graphing utility, we find the graph of $f (x) = \frac{e^{x}}{e^{x} - 4}$ is below the graph of $g ⁢ open parenthesis x close parenthesis equals 3$ on $open parenthesis minus ∞, ln ⁡ open parenthesis 4 close parenthesis close parenthesis union open parenthesis ln ⁡ open parenthesis 6 close parenthesis, ∞ close parenthesis$ , and they intersect at $x \approx 1.792 \approx \ln (6)$ .
Figure 7.71
Figure 7.72
A Sign Diagram for $r (x) = \frac{12 - 2 e^{x}}{e^{x} - 4}$
Checking $\frac{e^{x}}{e^{x} - 4} \leq 3$
As before, we start solving $t ⁢ e to the power 2 ⁢ t less than 4 ⁢ t$ by getting $0$ on one side of the inequality, $t e^{2 t} - 4 t < 0$ .
We set $r (t) = t e^{2 t} - 4 t$ and since there are no denominators, even-indexed radicals, or logs, the domain of $r$ is all real numbers.
Setting $r ⁢ open parenthesis t close parenthesis equals 0$ produces $t e^{2 t} - 4 t = 0$ . We factor to get $t (e^{2 t} - 4) = 0$ which gives $t equals 0$ or $e^{2 t} - 4 = 0$ .
To solve the latter, we isolate the exponential and take logs to get $2 ⁢ t equals ln ⁡ open parenthesis 4 close parenthesis$ , or $t = \frac{\ln (4)}{2}$ which simplifies to $t equals ln ⁡ open parenthesis 2 close parenthesis$ . (Can you see why?)
As in the previous example, we need to be careful about choosing test values. Since $ln ⁡ open parenthesis 1 close parenthesis equals 0$ , we choose $\ln (\frac{1}{2})$ , $\ln (\frac{3}{2})$ and $ln ⁡ open parenthesis 3 close parenthesis$ . Evaluating,⁶ we get
$\begin{matrix} r (\ln (\frac{1}{2})) & = & \ln (\frac{1}{2}) e^{2 \ln (\frac{1}{2})} - 4 \ln (\frac{1}{2}) \\ = & \ln (\frac{1}{2}) e^{\ln {(\frac{1}{2})}^{2}} - 4 \ln (\frac{1}{2}) & Power Rule \\ = & \ln (\frac{1}{2}) e^{\ln (\frac{1}{4})} - 4 \ln (\frac{1}{2}) \\ = & \frac{1}{4} \ln (\frac{1}{2}) - 4 \ln (\frac{1}{2}) = - \frac{15}{4} \ln (\frac{1}{2}) \end{matrix}$
Since $\frac{1}{2} < 1$ , $\ln (\frac{1}{2}) < 0$ and we get $r (\ln (\frac{1}{2}))$ is $(+)$ . Proceeding similarly, we find $r (\ln (\frac{3}{2})) < 0$ and $r ⁢ open parenthesis ln ⁡ open parenthesis 3 close parenthesis close parenthesis greater than 0$ . Our solution corresponds to $r ⁢ open parenthesis t close parenthesis less than 0$ which occurs on $open parenthesis 0, ln ⁡ open parenthesis 2 close parenthesis close parenthesis$ .
The graphing utility confirms that the graph of $f (t) = t e^{2 t}$ is below the graph of $g ⁢ open parenthesis t close parenthesis equals 4 ⁢ t$ on $open parenthesis 0, ln ⁡ open parenthesis 2 close parenthesis close parenthesis$ .⁷
Figure 7.73
Figure 7.74
A Sign Diagram for $r (t) = t e^{2 t} - 4 t$
Checking $t ⁢ e to the power 2 ⁢ t less than 4 ⁢ t$

We note here that while sign diagrams will always work for solving inequalities involving exponential functions, as we've seen previously, there are circumstances in which we can short-cut this method.

For example, consider number from Example 7.4.2 above: $2^{x^{2} - 3 x} - 16 \geq 0$ . Since the base $2 greater than 1$ , $log sub 2 ⁡ open parenthesis x close parenthesis$ is an increasing function meaning it preserves inequalities.

We can use this to our advantage in this case and eliminate the exponential from the inequality altogether:

$\begin{matrix} 2^{x^{2} - 3 x} - 16 & \geq & 0 \\ 2^{x^{2} - 3 x} & \geq & 16 \\ \log_{2} (2^{x^{2} - 3 x}) & \geq & \log_{2} (16) & f (x) = \log_{2} (x) is increasing so if b \geq a, \log_{2} (b) \geq \log_{2} (a) . \\ x^{2} - 3 x & \geq & 4 \end{matrix}$

Hence, we've reduced our given inequality to $x squared minus 3 ⁢ x greater than or equal to 4$ . As seen in Section 1.4, we can solve this inequality by completing the square, graphing, or a sign diagram, whichever strikes the reader's fancy.

Our next example is a follow-up to Example 7.1.3 in Section 7.1.

Example 3

Recall from Example 7.1.3 the temperature of coffee $T$ (in degrees Fahrenheit) $t$ minutes after it is served can be modeled by $T (t) = 70 + 90 e^{- 0.1 t}$ . When will the coffee be warmer than $100 degrees ⁢ F$ ?

We need to find when $T ⁢ open parenthesis t close parenthesis greater than 100$ , that is, we need to solve $70 + 90 e^{- 0.1 t} > 100$ .

To use a sign diagram, we need to get $0$ on one side of the inequality. Subtracting $100$ from both sides of $70 + 90 e^{- 0.1 t} > 100$ produces $90 e^{- 0.1 t} - 30 > 0$ .

Identifying $r (t) = 90 e^{- 0.1 t} - 30$ , we note from the context of the problem the domain of $r$ is $open bracket 0, ∞ close parenthesis$ , so to build the sign diagram, we proceed to find the zeros of $r$ .

Solving $90 e^{- 0.1 t} - 30 = 0$ results in $e^{- 0.1 t} = \frac{1}{3}$ so that $t = - 10 \ln (\frac{1}{3})$ which reduces to $t equals 10 ⁢ ln ⁡ open parenthesis 3 close parenthesis$ .

If we wish to avoid using the calculator to choose test values, we note that $f (x) = \ln (x)$ is increasing. As a result, since $1 less than 3$ , $0 equals ln ⁡ open parenthesis 1 close parenthesis less than ln ⁡ open parenthesis 3 close parenthesis$ which proves $10 ⁢ ln ⁡ open parenthesis 3 close parenthesis greater than 0$ . Hence, we may choose $t equals 0$ as a test value in $open bracket 0, 10 ⁢ ln ⁡ open parenthesis 3 close parenthesis close parenthesis$ . Since $3 less than 4$ , $ln ⁡ open parenthesis 3 close parenthesis less than ln ⁡ open parenthesis 4 close parenthesis$ , so $10 ⁢ ln ⁡ open parenthesis 3 close parenthesis less than 10 ⁢ ln ⁡ open parenthesis 4 close parenthesis$ . Hence, we may choose $10 ⁢ ln ⁡ open parenthesis 4 close parenthesis$ as test value for the interval $open parenthesis 10 ⁢ ln ⁡ open parenthesis 3 close parenthesis, ∞ close parenthesis$ .

We find $r ⁢ open parenthesis 0 close parenthesis greater than 0$ and $r ⁢ open parenthesis 10 ⁢ ln ⁡ open parenthesis 4 close parenthesis close parenthesis less than 0$ which gives the sign diagram below. We see $r ⁢ open parenthesis t close parenthesis greater than 0$ on $open bracket 0, 10 ⁢ ln ⁡ open parenthesis 3 close parenthesis close parenthesis$ .

We graph $y = T (t)$ from Example 7.1.3 below on the right along with with the horizontal line $y equals 100$ . We see the graph of $T$ is above the horizontal line to the left of the intersection point, which we leave to the reader to show is $open parenthesis 10 ⁢ ln ⁡ open parenthesis 3 close parenthesis, 100 close parenthesis$ .

Figure 7.75

Figure 7.76

y = T (t)

Hence, the coffee is warmer than $100 degrees$ F up to $10 ⁢ ln ⁡ open parenthesis 3 close parenthesis approximately equals 11$ minutes after it is served, or, said differently, it takes approximately 11 minutes for the coffee to cool to under $100 degrees$ F.

We note that, once again, we can short-cut the sign diagram in Example 7.4.3 to solve $70 + 90 e^{- 0.1 t} > 100$ . Since $\ln (x)$ is increasing, it preserves inequality. This means we can solve this inequality as follows.

$\begin{matrix} 70 + 90 e^{- 0.1 t} & > & 100 \\ 90 e^{- 0.1 t} & > & 30 \\ e^{- 0.1 t} > & \frac{1}{3} \\ \ln (e^{- 0.1 t}) & > & \ln (\frac{1}{3}) & f (x) = \ln (x) is increasing so if b \geq a, \ln (b) \geq \ln (a) . \\ - 0.1 t & > & - \ln (3) & \ln (\frac{1}{3}) = \ln (3^{- 1}) = - \ln (3) . \\ t & < & \frac{- \ln (3)}{- 0.1} = 10 \ln (3) \end{matrix}$

Since we are given $t greater than or equal to 0$ , we arrive at the same answer $0 less than or equal to t less than 10 ⁢ ln ⁡ open parenthesis 3 close parenthesis$ or $open bracket 0, 10 ⁢ ln ⁡ open parenthesis 3 close parenthesis close parenthesis$ .

Note the importance, once again, of having a base larger than $1$ so that the corresponding logarithmic function is increasing. We can still adapt this strategy to exponential functions whose base is less than $1$ , but we need to remember the corresponding logarithmic function is decreasing so it reverses inequalities.

Example 4

The function $f (x) = \frac{5 e^{x}}{e^{x} + 1}$ is one-to-one.

Find a formula for $f^{- 1} (x)$ .
Solve $\frac{5 e^{x}}{e^{x} + 1} = 4$ .

We start by writing $y = f (x)$ , and interchange the roles of $x$ and $y$ . To solve for $y$ , we first clear denominators and then isolate the exponential function.
$\begin{matrix} y & = & \frac{5 e^{x}}{e^{x} + 1} \\ x & = & \frac{5 e^{y}}{e^{y} + 1} & Switch x and y \\ x (e^{y} + 1) & = & 5 e^{y} \\ x e^{y} + x & = & 5 e^{y} \\ x & = & 5 e^{y} - x e^{y} \\ x & = & e^{y} (5 - x) \\ e^{y} & = & \frac{x}{5 - x} \\ \ln (e^{y}) & = & \ln (\frac{x}{5 - x}) \\ y & = & \ln (\frac{x}{5 - x}) \end{matrix}$
We claim $f^{- 1} (x) = \ln (\frac{x}{5 - x})$ . To verify this analytically, we would need to verify the compositions $(f^{- 1} \circ f) (x) = x$ for all $x$ in the domain of $f$ and that $(f \circ f^{- 1}) (x) = x$ for all $x$ in the domain of $f^{- 1}$ . We leave this, as well as a graphical check, to the reader in Exercise.
We recognize the equation $\frac{5 e^{x}}{e^{x} + 1} = 4$ as $f ⁢ open parenthesis x close parenthesis equals 4$ . Hence, our solution is $x = f^{- 1} (4) = \ln (\frac{4}{5 - 4}) = \ln (4)$ .
We can check this fairly quickly algebraically. Using $e^{\ln (4)} = 4$ , we find $\frac{5 e^{\ln (4)}}{e^{\ln (4)} + 1} = \frac{5 (4)}{4 + 1} = \frac{20}{5} = 4$ .

Our last example uses the tools of this section along with those developed in Section 6.3.

Example 5

Let $f (x) = 3 x e^{- x}$ .

Given that $f^{'} (x) = 3 e^{- x} - 3 x e^{- x}$ , find the open intervals over which $f$ is increasing and decreasing.
Find the local extrema.⁸
Given that $f^{''} (x) = 3 x e^{- x} - 6 e^{- x}$ , find the open intervals over which the graph of $f$ is concave up and concave down.
Locate the inflection points.
Check your answers using a graphing utility.

To determine where $f$ is increasing and decreasing, we need to make a sign diagram for $f^{'} (x)$ . Since the domain of $f^{'}$ is all real numbers, we just need to find the zeros of $f^{'}$ .
Solving $f^{'} (x) = 3 e^{- x} - 3 x e^{- x} = 0$ gives $3 e^{- x} (1 - x) = 0$ so $3 e^{- x} = 0$ , which has no solution, or $1 - x = 0$ so $x equals 1$ . We make a sign diagram for $f^{'} (x)$ below.
Figure 7.77
Figure 7.78
We get $f$ is increasing on $open parenthesis minus ∞, 1 close parenthesis$ and decreasing on $open parenthesis 1, ∞ close parenthesis$ .
Using the sign diagram for $f^{'} (x)$ , we see we have a local maximum at $(1, f (1)) = (1, 3 e^{- 1})$ .
Once again, the domain of $f^{''}$ is all real numbers, so our first step in constructing a sign diagram for $f^{''} (x)$ is to find the zeros.
Solving $f^{''} (x) = 3 x e^{- x} - 6 e^{- x} = 0$ gives $3 e^{- x} (x - 2) = 0$ so either $3 e^{- x} = 0$ , which has no solution, or $x - 2 = 0$ , so $x equals 2$ . We make the sign diagram for $f^{''} (x)$ below.
Figure 7.79
Figure 7.80
We see the graph of $f$ is concave up on $open parenthesis minus ∞, 2 close parenthesis$ and concave down on $open parenthesis 2, ∞ close parenthesis$ .
Since $f$ changes concavity at $x equals 2$ , we have an inflection point at $(2, f (2)) = (2, 6 e^{- 2})$ .
We check using desmos :
Figure 7.81

Note that the graph of $f (x) = 3 x e^{- x}$ produced by the graphing utility in Example 7.4.5 suggests the graph of $f$ has a horizontal asymptote as $x → ∞$ .

Indeed, it is the case that $lim x → ∞ f ⁢ open parenthesis x close parenthesis equals 0$ , however if we try to reason this analytically, we get another instance of an indeterminate form.⁹ As $x → ∞$ , $3 ⁢ x → ∞$ but $e^{- x} \to 0$ . Hence, as $x → ∞$ , we get the indeterminate form ` $∞ times 0$ .' Depending on how quickly the first factor approaches ` $∞$ ' and how quickly the second factor approaches ` $0$ ', we could end up with ` $∞$ ', ` $0$ ,' or some number in between.¹⁰

We'll explore more of this phenomenon in Exercise.¹¹ For now, we take it as true that exponential functions dominate polynomial functions so in the above indeterminate form, the factor $e^{- x}$ determines¹² the end behavior of $f$ , so $lim x → ∞ f ⁢ open parenthesis x close parenthesis equals 0$ .

Exercises

In Exercises -, solve the equation analytically.

$2^{4 x} = 8$
$3^{(x - 1)} = 27$
$5^{2 x - 1} = 125$
$4^{2 t} = \frac{1}{2}$
$8^{t} = \frac{1}{128}$
$2^{(t^{3} - t)} = 1$
$3^{7 x} = 81^{4 - 2 x}$
$9 \cdot 3^{7 x} = {(\frac{1}{9})}^{2 x}$
$3^{2 x} = 5$
$5^{- t} = 2$
$5^{t} = - 2$
$3^{(t - 1)} = 29$
${(1.005)}^{12 x} = 3$
$e^{- 5730 k} = \frac{1}{2}$
$2000 e^{0.1 t} = 4000$
$500 (1 - e^{2 t}) = 250$
$70 + 90 e^{- 0.1 t} = 75$
$30 - 6 e^{- 0.1 t} = 20$
$\frac{100 e^{x}}{e^{x} + 2} = 50$
$\frac{5000}{1 + 2 e^{- 3 t}} = 2500$
$\frac{150}{1 + 29 e^{- 0.8 t}} = 75$
$25 {(\frac{4}{5})}^{x} = 10$
$e^{2 x} = 2 e^{x}$
$7 e^{2 t} = 28 e^{- 6 t}$
$3^{(x - 1)} = 2^{x}$
$3^{(x - 1)} = {(\frac{1}{2})}^{(x + 5)}$
$7^{3 + 7 x} = 3^{4 - 2 x}$
$e^{2 t} - 3 e^{t} - 10 = 0$
$e^{2 t} = e^{t} + 6$
$4^{t} + 2^{t} = 12$
$e^{x} - 3 e^{- x} = 2$
$e^{x} + 15 e^{- x} = 8$
$3^{x} + 25 \cdot 3^{- x} = 10$

In Exercises -, solve the inequality analytically.

$e to the power x greater than 53$
$1000 ⁢ open parenthesis 1.005 close parenthesis to the power 12 ⁢ t greater than or equal to 3000$
$2^{(x^{3} - x)} < 1$
$25 {(\frac{4}{5})}^{x} \geq 10$
$\frac{150}{1 + 29 e^{- 0.8 t}} \leq 130$
$70 + 90 e^{- 0.1 t} \leq 75$
$e^{- x} - x e^{- x} \geq 0$
$(1 - e^{t}) t^{- 1} \leq 0$

In Exercises -, use a graphing utility to help you solve the equation or inequality.

$2^{x} = x^{2}$
$e^{t} = \ln (t) + 5$
$e^{\sqrt{x}} = x + 1$
$e^{- 2 t} - t e^{- t} \leq 0$
$3^{(x - 1)} < 2^{x}$
$e^{t} < t^{3} - t$

In Exercises -, find the domain of the function.

$T (x) = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$
$C (x) = \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}}$
$s (t) = \sqrt{e^{2 t} - 3}$
$c (t) = \sqrt[3]{e^{2 t} - 3}$
$L (x) = \log (3 - e^{x})$
$ℓ (x) = \ln (\frac{e^{2 x}}{e^{x} - 2})$
Since $f (x) = \ln (x)$ is a strictly increasing function, if $0 less than a less than b$ then $\ln (a) < \ln (b)$ . Use this fact to solve the inequality $e^{(3 x - 1)} > 6$ without a sign diagram. Use this technique to solve the inequalities in Exercises -. (NOTE: Isolate the exponential function first!)
Compute the inverse of $f (x) = \frac{e^{x} - e^{- x}}{2}$ . State the domain and range of both $f$ and $f^{- 1}$ .
In Example 7.4.4, we found that the inverse of $f (x) = \frac{5 e^{x}}{e^{x} + 1}$ was $f^{- 1} (x) = \ln (\frac{x}{5 - x})$ but we left a few loose ends for you to tie up.
1. Algebraically check our answer by verifying: $(f^{- 1} \circ f) (x) = x$ for all $x$ in the domain of $f$ and that $(f \circ f^{- 1}) (x) = x$ for all $x$ in the domain of $f^{- 1}$ .
2. Find the range of $f$ by finding the domain of $f^{- 1}$ .
3. With help of a graphing utility, graph $y = f (x)$ , $y = f^{- 1} (x)$ and $y = x$ on the same set of axes. How does this help to verify our answer?
4. Let $g (x) = \frac{5 x}{x + 1}$ and $h (x) = e^{x}$ . Show that $f equals g ∘ h$ and that ${(g \circ h)}^{- 1} = h^{- 1} \circ g^{- 1}$ .
  NOTE: We know this is true in general by Exercise in Section 5.6, but it's nice to see a specific example of the property.
1. With the help of your classmates, numerically and graphically investigate $lim x → ∞$ $\frac{x^{p}}{e^{x}}$ for various real number powers, $p$ .
2. What does part suggest about the relative growth rates of powers of $x$ as opposed to $e^{x}$ ?
3. For each power $p$ you investigated in part, solve the inequality: $\frac{x^{p}}{e^{x}} < \frac{1}{x}$ .
4. Use your results from part to show that for each real number $p$ you investigated in part, there is a real number $M$ so that if $x > M$ , $0 < \frac{x^{p}}{e^{x}} < \frac{1}{x}$ .
  Since $lim x → ∞$ $\frac{1}{x} = 0$ , what do you conclude about $lim x → ∞$ $\frac{x^{p}}{e^{x}}$ ?
  (This Exercise foreshadows the celebrated Squeeze Theorem, Theorem 10.2 which we'll formally introduce in Section 10.1.)

In Exercises - a function $f$ along with its derivatives $f^{'}$ and $f^{''}$ are given.

Find the $x$ - and $y$ -intercepts of the graph of each function, if any.
Use limits to determine the end behavior.
Use $f^{'}$ to determine the open intervals over which $f$ is increasing or decreasing.
Determine the local extrema, if any.
Use $f^{''}$ to determine the open intervals over which the graph of $f$ is concave up or concave down.
Determine the inflection points of the graph, if any.
$f (x) = \frac{5}{1 + e^{- x}}$ , $f^{'} (x) = \frac{5 e^{- x}}{{(1 + e^{- x})}^{2}}$ , $f^{''} (x) = \frac{5 e^{- x} (e^{- x} - 1)}{{(1 + e^{- x})}^{3}}$
$f (x) = e^{- x} - e^{- 2 x}$ , $f^{'} (x) = 2 e^{- 2 x} - e^{- x}$ , $f^{''} (x) = e^{- x} - 4 e^{- 2 x}$

Answers

$x = \frac{3}{4}$
$x equals 4$
$x equals 2$
$t = - \frac{1}{4}$
$t = - \frac{7}{3}$
$t = - 1, 0, 1$
$x = \frac{16}{15}$
$x = - \frac{2}{11}$
$x = \frac{\ln (5)}{2 \ln (3)}$
$t = - \frac{\ln (2)}{\ln (5)}$
No solution.
$t = \frac{\ln (29) + \ln (3)}{\ln (3)}$
$x = \frac{\ln (3)}{12 \ln (1.005)}$
$k = \frac{\ln (\frac{1}{2})}{- 5730} = \frac{\ln (2)}{5730}$
$t = \frac{\ln (2)}{0.1} = 10 \ln (2)$
$t = \frac{1}{2} \ln (\frac{1}{2}) = - \frac{1}{2} \ln (2)$
$t = \frac{\ln (\frac{1}{18})}{- 0.1} = 10 \ln (18)$
$t = - 10 \ln (\frac{5}{3}) = 10 \ln (\frac{3}{5})$
$x equals ln ⁡ open parenthesis 2 close parenthesis$
$t = \frac{1}{3} \ln (2)$
$t = \frac{\ln (\frac{1}{29})}{- 0.8} = \frac{5}{4} \ln (29)$
$x = \frac{\ln (\frac{2}{5})}{\ln (\frac{4}{5})} = \frac{\ln (2) - \ln (5)}{\ln (4) - \ln (5)}$
$x equals ln ⁡ open parenthesis 2 close parenthesis$
$t = - \frac{1}{8} \ln (\frac{1}{4}) = \frac{1}{4} \ln (2)$
$x = \frac{\ln (3)}{\ln (3) - \ln (2)}$
$x = \frac{\ln (3) + 5 \ln (\frac{1}{2})}{\ln (3) - \ln (\frac{1}{2})} = \frac{\ln (3) - 5 \ln (2)}{\ln (3) + \ln (2)}$
$x = \frac{4 \ln (3) - 3 \ln (7)}{7 \ln (7) + 2 \ln (3)}$
$t equals ln ⁡ open parenthesis 5 close parenthesis$
$t equals ln ⁡ open parenthesis 3 close parenthesis$
$t = \frac{\ln (3)}{\ln (2)}$
$x equals ln ⁡ open parenthesis 3 close parenthesis$
$x equals ln ⁡ open parenthesis 3 close parenthesis$ , $ln ⁡ open parenthesis 5 close parenthesis$
$x = \frac{\ln (5)}{\ln (3)}$
$open parenthesis ln ⁡ open parenthesis 53 close parenthesis, ∞ close parenthesis$
$[\frac{\ln (3)}{12 \ln (1.005)}, \infty)$
$(- \infty, - 1) \cup (0, 1)$
$(- \infty, \frac{\ln (\frac{2}{5})}{\ln (\frac{4}{5})}] = (- \infty, \frac{\ln (2) - \ln (5)}{\ln (4) - \ln (5)}]$
$(- \infty, \frac{\ln (\frac{2}{377})}{- 0.8}] = (- \infty, \frac{5}{4} \ln (\frac{377}{2})]$
$[\frac{\ln (\frac{1}{18})}{- 0.1}, \infty) = [10 \ln (18), \infty)$
$open parenthesis minus ∞, 1 close bracket$
$open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$
$x \approx - 0.76666, x = 2, x = 4$
$x \approx 0.01866, x \approx 1.7115$
$x equals 0$
$approximately equals open bracket 0.567, ∞ close parenthesis$
$\approx (- \infty, 2.7095)$
$approximately equals open parenthesis 2.3217, 4.3717 close parenthesis$
$open parenthesis minus ∞, ∞ close parenthesis$
$open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$
$(\frac{1}{2} \ln (3), \infty)$
$open parenthesis minus ∞, ∞ close parenthesis$
$open parenthesis minus ∞, ln ⁡ open parenthesis 3 close parenthesis close parenthesis$
$open parenthesis ln ⁡ open parenthesis 2 close parenthesis, ∞ close parenthesis$
$x > \frac{1}{3} (\ln (6) + 1)$ , so $(\frac{1}{3} (\ln (6) + 1), \infty)$
$f^{- 1} = \ln (x + \sqrt{x^{2} + 1})$ . Both $f$ and $f^{- 1}$ have domain $open parenthesis minus ∞, ∞ close parenthesis$ and range $open parenthesis minus ∞, ∞ close parenthesis$ .
- There are no $x$ -intercepts; the $y$ -intercept is $open parenthesis 0, 5 close parenthesis$ .
- $lim_{x \to - \infty} f (x) = 0$ and $lim x → ∞ f ⁢ open parenthesis x close parenthesis equals 5$ ; we have two horizontal asymptotes: $y equals 0$ and $y equals 5$ .
- $f$ is always increasing: $open parenthesis minus ∞, ∞ close parenthesis$ .
- There are no local extrema.
- The graph of $f$ is concave up on $open parenthesis minus ∞, 0 close parenthesis$ and concave down on $open parenthesis 0, ∞ close parenthesis$ .
- The inflection point is $open parenthesis 0, 5 close parenthesis$ .
- The $x$ - and $y$ -intercept is $open parenthesis 0, 0 close parenthesis$ .
- $lim_{x \to - \infty} f (x) = - \infty$ and $lim x → ∞ f ⁢ open parenthesis x close parenthesis equals 0$ ; we have a horizontal asymptote $y equals 0$ .
- $f$ is increasing on $open parenthesis minus ∞, ln ⁡ open parenthesis 2 close parenthesis close parenthesis$ and decreasing on $open parenthesis ln ⁡ open parenthesis 2 close parenthesis, ∞ close parenthesis$ .
- There is a local (absolute) maximum at $(\ln (2), \frac{1}{4})$ .
- The graph of $f$ is concave up on $open parenthesis ln ⁡ open parenthesis 4 close parenthesis, ∞ close parenthesis$ and concave down on $open parenthesis minus ∞, ln ⁡ open parenthesis 4 close parenthesis close parenthesis$ .
- The inflection point is $(\ln (4), \frac{3}{16})$ .

This is also the `if' part of the statement $\log_{b} (u) = \log_{b} (w)$ if and only if $u = w$ in Theorem 7.5. ↩
Please resist the temptation to divide both sides by ` $\ln$ ' instead of $ln ⁡ open parenthesis 2 close parenthesis$ . Just like it wouldn't make sense to divide both sides by the square root symbol ` $\sqrt{}$ ' when solving $x \sqrt{2} = 5$ , it makes no sense to divide by ` $\ln$ '. ↩
Why not? ↩
This is because the base of $\ln (x)$ is $e greater than 1$ . If the base $b$ were in the interval $0 less than b less than 1$ , then $\log_{b} (x)$ would decreasing. ↩
We could, of course, use the calculator, but what fun would that be? ↩
A calculator can be used at this point. As usual, we proceed without apologies, with the analytical method. ↩
Note: $ln ⁡ open parenthesis 2 close parenthesis approximately equals 0.693$ . ↩
Recall this means the local maximums and minimums. ↩
Previously, we've seen the indeterminate form ` $\frac{0}{0}$ .' ↩
It is important here to understand that the factor $e^{- x}$ which results in the ` $0$ ' in the indeterminate form ` $∞ times 0$ ' is approaching $0$ and is not actually equal to $0$ . An incorrect reasoning that the form $∞ times 0 → 0$ in this case is that `anything times $0$ is $0$ .' Again, we'll have more to say about this in the Exercises. ↩
With formal proofs found in Calculus … ↩
Less formally, the factor $e^{- x} \to 0$ `faster' than the factor $3 ⁢ x → ∞$ which forces the product $3 x e^{- x} \to 0$ . ↩

Adapted from Precalculus, Preliminary 4th Edition (integrated calculus), by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.