6.4 Logarithmic Equations and Inequalities

In Section we solved equations and inequalities involving exponential functions using one of two basic strategies. We now turn our attention to equations and inequalities involving logarithmic functions, and not surprisingly, there are two basic strategies to choose from. For example, suppose we wish to solve $log sub 2 ⁡ open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis 5 close parenthesis$ . Theorem tells us that the only solution to this equation is $x equals 5$ . Now suppose we wish to solve $log sub 2 ⁡ open parenthesis x close parenthesis equals 3$ . If we want to use Theorem, we need to rewrite $3$ as a logarithm base $2$ . We can use Theorem to do just that: $3 = \log_{2} (2^{3}) = \log_{2} (8)$ . Our equation then becomes $log sub 2 ⁡ open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis 8 close parenthesis$ so that $x equals 8$ . However, we could have arrived at the same answer, in fewer steps, by using Theorem to rewrite the equation $log sub 2 ⁡ open parenthesis x close parenthesis equals 3$ as $2 cubed equals x$ , or $x equals 8$ . We summarize the two common ways to solve log equations below.

Steps for Solving an Equation involving Logarithmic Functions

Isolate the logarithmic function.
1. If convenient, express both sides as logs with the same base and equate the arguments of the log functions.
2. Otherwise, rewrite the log equation as an exponential equation.

Example 1

Solve the following equations. Check your solutions graphically using a calculator.

$\log_{117} (1 - 3 x) = \log_{117} (x^{2} - 3)$
$2 - \ln (x - 3) = 1$
$\log_{6} (x + 4) + \log_{6} (3 - x) = 1$
$\log_{7} (1 - 2 x) = 1 - \log_{7} (3 - x)$
$\log_{2} (x + 3) = \log_{2} (6 - x) + 3$
$1 + 2 \log_{4} (x + 1) = 2 \log_{2} (x)$

Since we have the same base on both sides of the equation $\log_{117} (1 - 3 x) = \log_{117} (x^{2} - 3)$ , we equate what's inside the logs to get $1 - 3 x = x^{2} - 3$ . Solving $x^{2} + 3 x - 4 = 0$ gives $x = - 4$ and $x equals 1$ . To check these answers using the calculator, we make use of the change of base formula and graph $f (x) = \frac{\ln (1 - 3 x)}{\ln (117)}$ and $g (x) = \frac{\ln (x^{2} - 3)}{\ln (117)}$ and we see they intersect only at $x = - 4$ . To see what happened to the solution $x equals 1$ , we substitute it into our original equation to obtain $\log_{117} (- 2) = \log_{117} (- 2)$ . While these expressions look identical, neither is a real number,¹ which means $x equals 1$ is not in the domain of the original equation, and is not a solution.
Our first objective in solving $2 - \ln (x - 3) = 1$ is to isolate the logarithm. We get $\ln (x - 3) = 1$ , which, as an exponential equation, is $e^{1} = x - 3$ . We get our solution $x = e + 3$ . On the calculator, we see the graph of $f (x) = 2 - \ln (x - 3)$ intersects the graph of $g ⁢ open parenthesis x close parenthesis equals 1$ at $x = e + 3 \approx 5.718$ .
Figure 6.56
Figure 6.57
$y = f (x) = \log_{117} (1 - 3 x)$ and
$y = f (x) = 2 - \ln (x - 3)$ and
$y = g (x) = \log_{117} (x^{2} - 3)$
$y = g (x) = 1$
We can start solving $\log_{6} (x + 4) + \log_{6} (3 - x) = 1$ by using the Product Rule for logarithms to rewrite the equation as $\log_{6} [(x + 4) (3 - x)] = 1$ . Rewriting this as an exponential equation, we get $6^{1} = (x + 4) (3 - x)$ . This reduces to $x^{2} + x - 6 = 0$ , which gives $x = - 3$ and $x equals 2$ . Graphing $y = f (x) = \frac{\ln (x + 4)}{\ln (6)} + \frac{\ln (3 - x)}{\ln (6)}$ and $y = g (x) = 1$ , we see they intersect twice, at $x = - 3$ and $x equals 2$ .
Figure 6.58
Figure 6.59
$y = f (x) = \log_{6} (x + 4) + \log_{6} (3 - x)$ and $y = g (x) = 1$
Taking a cue from the previous problem, we begin solving $\log_{7} (1 - 2 x) = 1 - \log_{7} (3 - x)$ by first collecting the logarithms on the same side, $\log_{7} (1 - 2 x) + \log_{7} (3 - x) = 1$ , and then using the Product Rule to get $\log_{7} [(1 - 2 x) (3 - x)] = 1$ . Rewriting this as an exponential equation gives $7^{1} = (1 - 2 x) (3 - x)$ which gives the quadratic equation $2 x^{2} - 7 x - 4 = 0$ . Solving, we find $x = - \frac{1}{2}$ and $x equals 4$ . Graphing, we find $y = f (x) = \frac{\ln (1 - 2 x)}{\ln (7)}$ and $y = g (x) = 1 - \frac{\ln (3 - x)}{\ln (7)}$ intersect only at $x = - \frac{1}{2}$ . Checking $x equals 4$ in the original equation produces $\log_{7} (- 7) = 1 - \log_{7} (- 1)$ , which is a clear domain violation.
Starting with $\log_{2} (x + 3) = \log_{2} (6 - x) + 3$ , we gather the logarithms to one side and get $\log_{2} (x + 3) - \log_{2} (6 - x) = 3$ . We then use the Quotient Rule and convert to an exponential equation
$\log_{2} (\frac{x + 3}{6 - x}) = 3 \Leftrightarrow 2^{3} = \frac{x + 3}{6 - x}$
This reduces to the linear equation $8 (6 - x) = x + 3$ , which gives us $x equals 5$ . When we graph $f (x) = \frac{\ln (x + 3)}{\ln (2)}$ and $g (x) = \frac{\ln (6 - x)}{\ln (2)} + 3$ , we find they intersect at $x equals 5$ .
Figure 6.60
Figure 6.61
$y = f (x) = \log_{7} (1 - 2 x)$ and
$y = f (x) = \log_{2} (x + 3)$ and
$y = g (x) = 1 - \log_{7} (3 - x)$
$y = g (x) = \log_{2} (6 - x) + 3$
Starting with $1 + 2 \log_{4} (x + 1) = 2 \log_{2} (x)$ , we gather the logs to one side to get the equation $1 = 2 \log_{2} (x) - 2 \log_{4} (x + 1)$ . Before we can combine the logarithms, however, we need a common base. Since $4$ is a power of $2$ , we use change of base to convert
$\log_{4} (x + 1) = \frac{\log_{2} (x + 1)}{\log_{2} (4)} = \frac{1}{2} \log_{2} (x + 1)$
Hence, our original equation becomes
$\begin{matrix} 1 & = & 2 \log_{2} (x) - 2 (\frac{1}{2} \log_{2} (x + 1)) \\ 1 & = & 2 \log_{2} (x) - \log_{2} (x + 1) \\ 1 & = & \log_{2} (x^{2}) - \log_{2} (x + 1) & Power Rule \\ 1 & = & \log_{2} (\frac{x^{2}}{x + 1}) & Quotient Rule \end{matrix}$
Rewriting this in exponential form, we get $\frac{x^{2}}{x + 1} = 2$ or $x^{2} - 2 x - 2 = 0$ . Using the quadratic formula, we get $x = 1 \pm \sqrt{3}$ . Graphing $f (x) = 1 + \frac{2 \ln (x + 1)}{\ln (4)}$ and $g (x) = \frac{2 \ln (x)}{\ln (2)}$ , we see the graphs intersect only at $x = 1 + \sqrt{3} \approx 2.732$ . The solution $x = 1 - \sqrt{3} < 0$ , which means if substituted into the original equation, the term $2 \log_{2} (1 - \sqrt{3})$ is undefined.
Figure 6.62
$y = f (x) = 1 + 2 \log_{4} (x + 1)$ and $y = g (x) = 2 \log_{2} (x)$

If nothing else, Example Example 1 demonstrates the importance of checking for extraneous solutions² when solving equations involving logarithms. Even though we checked our answers graphically, extraneous solutions are easy to spot - any supposed solution which causes a negative number inside a logarithm needs to be discarded. As with the equations in Example, much can be learned from checking all of the answers in Example Example 1 analytically. We leave this to the reader and turn our attention to inequalities involving logarithmic functions. Since logarithmic functions are continuous on their domains, we can use sign diagrams.

Example 2

Solve the following inequalities. Check your answer graphically using a calculator.

$\frac{1}{\ln (x) + 1} \leq 1$
$open parenthesis log sub 2 ⁡ open parenthesis x close parenthesis close parenthesis squared less than 2 ⁢ log sub 2 ⁡ open parenthesis x close parenthesis plus 3$
$x ⁢ log ⁡ open parenthesis x plus 1 close parenthesis greater than or equal to x$

We start solving $\frac{1}{\ln (x) + 1} \leq 1$ by getting $0$ on one side of the inequality: $\frac{1}{\ln (x) + 1} - 1 \leq 0$ . Getting a common denominator yields $\frac{1}{\ln (x) + 1} - \frac{\ln (x) + 1}{\ln (x) + 1} \leq 0$ which reduces to $\frac{- \ln (x)}{\ln (x) + 1} \leq 0$ , or $\frac{\ln (x)}{\ln (x) + 1} \geq 0$ . We define $r (x) = \frac{\ln (x)}{\ln (x) + 1}$ and set about finding the domain and the zeros of $r$ . Due to the appearance of the term $\ln (x)$ , we require $x greater than 0$ . In order to keep the denominator away from zero, we solve $\ln (x) + 1 = 0$ so $\ln (x) = - 1$ , so $x = e^{- 1} = \frac{1}{e}$ . Hence, the domain of $r$ is $(0, \frac{1}{e}) \cup (\frac{1}{e}, \infty)$ . To find the zeros of $r$ , we set $r (x) = \frac{\ln (x)}{\ln (x) + 1} = 0$ so that $ln ⁡ open parenthesis x close parenthesis equals 0$ , and we find $x = e^{0} = 1$ . In order to determine test values for $r$ without resorting to the calculator, we need to find numbers between $0$ , $\frac{1}{e}$ , and $1$ which have a base of $e$ . Since $e approximately equals 2.718 greater than 1$ , $0 < \frac{1}{e^{2}} < \frac{1}{e} < \frac{1}{\sqrt{e}} < 1 < e$ . To determine the sign of $r (\frac{1}{e^{2}})$ , we use the fact that $\ln (\frac{1}{e^{2}}) = \ln (e^{- 2}) = - 2$ , and find $r (\frac{1}{e^{2}}) = \frac{- 2}{- 2 + 1} = 2$ , which is $(+)$ . The rest of the test values are determined similarly. From our sign diagram, we find the solution to be $(0, \frac{1}{e}) \cup [1, \infty)$ . Graphing $f (x) = \frac{1}{\ln (x) + 1}$ and $g ⁢ open parenthesis x close parenthesis equals 1$ , we see the graph of $f$ is below the graph of $g$ on the solution intervals, and that the graphs intersect at $x equals 1$ .
Figure 6.63
Figure 6.64
$y = f (x) = \frac{1}{\ln (x) + 1}$ and $y = g (x) = 1$
Moving all of the nonzero terms of $open parenthesis log sub 2 ⁡ open parenthesis x close parenthesis close parenthesis squared less than 2 ⁢ log sub 2 ⁡ open parenthesis x close parenthesis plus 3$ to one side of the inequality, we have ${(\log_{2} (x))}^{2} - 2 \log_{2} (x) - 3 < 0$ . Defining $r (x) = {(\log_{2} (x))}^{2} - 2 \log_{2} (x) - 3$ , we get the domain of $r$ is $open parenthesis 0, ∞ close parenthesis$ , due to the presence of the logarithm. To find the zeros of $r$ , we set $r (x) = {(\log_{2} (x))}^{2} - 2 \log_{2} (x) - 3 = 0$ which results in a `quadratic in disguise.' We set $u equals log sub 2 ⁡ open parenthesis x close parenthesis$ so our equation becomes $u^{2} - 2 u - 3 = 0$ which gives us $u = - 1$ and $u equals 3$ . Since $u equals log sub 2 ⁡ open parenthesis x close parenthesis$ , we get $\log_{2} (x) = - 1$ , which gives us $x = 2^{- 1} = \frac{1}{2}$ , and $log sub 2 ⁡ open parenthesis x close parenthesis equals 3$ , which yields $x = 2^{3} = 8$ . We use test values which are powers of $2$ : $0 < \frac{1}{4} < \frac{1}{2} < 1 < 8 < 16$ , and from our sign diagram, we see $r ⁢ open parenthesis x close parenthesis less than 0$ on $(\frac{1}{2}, 8)$ . Geometrically, we see the graph of $f (x) = {(\frac{\ln (x)}{\ln (2)})}^{2}$ is below the graph of $y = g (x) = \frac{2 \ln (x)}{\ln (2)} + 3$ on the solution interval.
Figure 6.65
Figure 6.66
$y = f (x) = {(\log_{2} (x))}^{2}$ and $y = g (x) = 2 \log_{2} (x) + 3$
We begin to solve $x ⁢ log ⁡ open parenthesis x plus 1 close parenthesis greater than or equal to x$ by subtracting $x$ from both sides to get $x \log (x + 1) - x \geq 0$ . We define $r (x) = x \log (x + 1) - x$ and due to the presence of the logarithm, we require $x plus 1 greater than 0$ , or $x greater than minus 1$ . To find the zeros of $r$ , we set $r (x) = x \log (x + 1) - x = 0$ . Factoring, we get $x (\log (x + 1) - 1) = 0$ , which gives $x equals 0$ or $\log (x + 1) - 1 = 0$ . The latter gives $\log (x + 1) = 1$ , or $x + 1 = 10^{1}$ , which admits $x equals 9$ . We select test values $x$ so that $x plus 1$ is a power of $10$ , and we obtain $- 1 < - 0.9 < 0 < \sqrt{10} - 1 < 9 < 99$ . Our sign diagram gives the solution to be $open parenthesis minus 1, 0 close bracket union open bracket 9, ∞ close parenthesis$ . The calculator indicates the graph of $y = f (x) = x \log (x + 1)$ is above $y = g (x) = x$ on the solution intervals, and the graphs intersect at $x equals 0$ and $x equals 9$ .
Figure 6.67
Figure 6.68
$y = f (x) = x \log (x + 1)$ and $y = g (x) = x$

Our next example revisits the concept of pH first seen in Exercise in Section.

Example 3

In order to successfully breed Ippizuti fish the pH of a freshwater tank must be at least 7.8 but can be no more than 8.5. Determine the corresponding range of hydrogen ion concentration, and check your answer using a calculator.

Recall from Exercise in Section that $pH = - \log [H^{+}]$ where $[H^{+}]$ is the hydrogen ion concentration in moles per liter. We require $7.8 \leq - \log [H^{+}] \leq 8.5$ or $- 7.8 \geq \log [H^{+}] \geq - 8.5$ . To solve this compound inequality we solve $- 7.8 \geq \log [H^{+}]$ and $\log [H^{+}] \geq - 8.5$ and take the intersection of the solution sets.³ The former inequality yields $0 < [H^{+}] \leq 10^{- 7.8}$ and the latter yields $[H^{+}] \geq 10^{- 8.5}$ . Taking the intersection gives us our final answer $10^{- 8.5} \leq [H^{+}] \leq 10^{- 7.8}$ . (Your Chemistry professor may want the answer written as $3.16 \times 10^{- 9} \leq [H^{+}] \leq 1.58 \times 10^{- 8}$ .) After carefully adjusting the viewing window on the graphing calculator we see that the graph of $f (x) = - \log (x)$ lies between the lines $y equals 7.8$ and $y equals 8.5$ on the interval $[3.16 \times 10^{- 9}, 1.58 \times 10^{- 8}]$ .

Figure 6.69

Figure 6.70

The graphs of $y = f (x) = - \log (x)$ , $y equals 7.8$ and $y equals 8.5$

We close this section by finding an inverse of a one-to-one function which involves logarithms.

Example 4

The function $f (x) = \frac{\log (x)}{1 - \log (x)}$ is one-to-one. Find a formula for $f^{- 1} (x)$ and check your answer graphically using your calculator.

We first write $y = f (x)$ then interchange the $x$ and $y$ and solve for $y$ .

$\begin{matrix} y & = & f (x) \\ y & = & \frac{\log (x)}{1 - \log (x)} \\ x & = & \frac{\log (y)}{1 - \log (y)} & Interchange x and y . \\ x (1 - \log (y)) & = & \log (y) \\ x - x \log (y) & = & \log (y) \\ x & = & x \log (y) + \log (y) \\ x & = & (x + 1) \log (y) \\ \frac{x}{x + 1} & = & \log (y) \\ y & = & 10^{\frac{x}{x + 1}} & Rewrite as an exponential equation. \end{matrix}$

We have $f^{- 1} (x) = 10^{\frac{x}{x + 1}}$ . Graphing $f$ and $f^{- 1}$ on the same viewing window yields

Figure 6.71

$y = f (x) = \frac{\log (x)}{1 - \log (x)}$ and $y = g (x) = 10^{\frac{x}{x + 1}}$

Exercises

In Exercises -, solve the equation analytically.

$\log (3 x - 1) = \log (4 - x)$
$log sub 2 ⁡ open parenthesis x cubed close parenthesis equals log sub 2 ⁡ open parenthesis x close parenthesis$
$\ln (8 - x^{2}) = \ln (2 - x)$
$\log_{5} (18 - x^{2}) = \log_{5} (6 - x)$
$\log_{3} (7 - 2 x) = 2$
$\log_{\frac{1}{2}} (2 x - 1) = - 3$
$\ln (x^{2} - 99) = 0$
$\log (x^{2} - 3 x) = 1$
$\log_{125} (\frac{3 x - 2}{2 x + 3}) = \frac{1}{3}$
$\log (\frac{x}{10^{- 3}}) = 4.7$
$- \log (x) = 5.4$
$10 \log (\frac{x}{10^{- 12}}) = 150$
$6 - 3 \log_{5} (2 x) = 0$
$3 \ln (x) - 2 = 1 - \ln (x)$
$\log_{3} (x - 4) + \log_{3} (x + 4) = 2$
$\log_{5} (2 x + 1) + \log_{5} (x + 2) = 1$
$\log_{169} (3 x + 7) - \log_{169} (5 x - 9) = \frac{1}{2}$
$\ln (x + 1) - \ln (x) = 3$
$2 \log_{7} (x) = \log_{7} (2) + \log_{7} (x + 12)$
$\log (x) - \log (2) = \log (x + 8) - \log (x + 2)$
$\log_{3} (x) = \log_{\frac{1}{3}} (x) + 8$
$ln ⁡ open parenthesis ln ⁡ open parenthesis x close parenthesis close parenthesis equals 3$
${(\log (x))}^{2} = 2 \log (x) + 15$
$\ln (x^{2}) = {(\ln (x))}^{2}$
$\frac{1 - \ln (x)}{x^{2}} < 0$
$x ⁢ ln ⁡ open parenthesis x close parenthesis minus x greater than 0$
$10 \log (\frac{x}{10^{- 12}}) \geq 90$
$5.6 \leq \log (\frac{x}{10^{- 3}}) \leq 7.1$
$2.3 less than minus log ⁡ open parenthesis x close parenthesis less than 5.4$
$\ln (x^{2}) \leq {(\ln (x))}^{2}$
$\ln (x) = e^{- x}$
$\ln (x) = \sqrt[4]{x}$
$ln ⁡ open parenthesis x squared plus 1 close parenthesis greater than or equal to 5$
$\ln (- 2 x^{3} - x^{2} + 13 x - 6) < 0$
Since $f (x) = e^{x}$ is a strictly increasing function, if $a < b$ then $e^{a} < e^{b}$ . Use this fact to solve the inequality $ln ⁡ open parenthesis 2 ⁢ x plus 1 close parenthesis less than 3$ without a sign diagram. Use this technique to solve the inequalities in Exercises -. (Compare this to Exercise in Section.)
Solve $\ln (3 - y) - \ln (y) = 2 x + \ln (5)$ for $y$ .
In Example Example 4 we found the inverse of $f (x) = \frac{\log (x)}{1 - \log (x)}$ to be $f^{- 1} (x) = 10^{\frac{x}{x + 1}}$ .
1. Show that $(f^{- 1} \circ f) (x) = x$ 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. Let $g (x) = \frac{x}{1 - x}$ and $h (x) = \log (x)$ . Show that $f equals g ∘ h$ and ${(g \circ h)}^{- 1} = h^{- 1} \circ g^{- 1}$ . (We know this is true in general by Exercise in Section, but it's nice to see a specific example of the property.)
Let $f (x) = \frac{1}{2} \ln (\frac{1 + x}{1 - x})$ . Compute $f^{- 1} (x)$ and find its domain and range.
Explain the equation in Exercise and the inequality in Exercise above in terms of the Richter scale for earthquake magnitude. (See Exercise in Section.)
Explain the equation in Exercise and the inequality in Exercise above in terms of sound intensity level as measured in decibels. (See Exercise in Section.)
Explain the equation in Exercise and the inequality in Exercise above in terms of the pH of a solution. (See Exercise in Section.)
With the help of your classmates, solve the inequality $\sqrt[n]{x} > \ln (x)$ for a variety of natural numbers $n$ . What might you conjecture about the “speed” at which $f (x) = \ln (x)$ grows versus any principal $n^{th}$ root function?

In Exercises -, solve the inequality analytically.

In Exercises -, use your calculator to help you solve the equation or inequality.

Answers

$x = \frac{5}{4}$
$x equals 1$
$x = - 2$
$x = - 3, 4$
$x = - 1$
$x = \frac{9}{2}$
$x = \pm 10$
$x = - 2, 5$
$x = - \frac{17}{7}$
$x = 10^{1.7}$
$x = 10^{- 5.4}$
$x equals 10 cubed$
$x = \frac{25}{2}$
$x = e^{3 / 4}$
$x equals 5$
$x = \frac{1}{2}$
$x equals 2$
$x = \frac{1}{e^{3} - 1}$
$x equals 6$
$x equals 4$
$x equals 81$
$x = e^{e^{3}}$
$x = 10^{- 3}, 10^{5}$
$x = 1, x = e^{2}$
$open parenthesis e, ∞ close parenthesis$
$open parenthesis e, ∞ close parenthesis$
$[10^{- 3}, \infty)$
$[10^{2.6}, 10^{4.1}]$
$(10^{- 5.4}, 10^{- 2.3})$
$open parenthesis 0, 1 close bracket union open bracket e squared, ∞ close parenthesis$
$x approximately equals 1.3098$
$x \approx 4.177, x \approx 5503.665$
$\approx (- \infty, - 12.1414) \cup (12.1414, \infty)$
$\approx (- 3.0281, - 3) \cup (0.5, 0.5991) \cup (1.9299, 2)$
$- \frac{1}{2} < x < \frac{e^{3} - 1}{2}$
$y = \frac{3}{5 e^{2 x} + 1}$
$f^{- 1} (x) = \frac{e^{2 x} - 1}{e^{2 x} + 1} = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$ . (To see why we rewrite this in this form, see Exercise in Section.) The domain of $f^{- 1}$ is $open parenthesis minus ∞, ∞ close parenthesis$ and its range is the same as the domain of $f$ , namely $open parenthesis minus 1, 1 close parenthesis$ .

Adapted from Precalculus, 3rd corrected edition, 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.