5.6 Solving Equations Involving Logarithmic Functions

Vanessa Coffelt

5.6 Solving Equations Involving Logarithmic Functions

In Section 5.5 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, per Theorem 5.7, the only solution to $\log_{2}(x) = \log_{2}(5)$ is $x=5$ . Now consider $\log_{2}(x) = 3$ . To use Theorem 5.7, we need to rewrite $3$ as a logarithm base $2$ . Theorem 5.6 gives us $3 = \log_{2}\left(2^{3}\right) = \log_{2}(8)$ . Hence, $\log_{2}(x) = 3$ is equivalent to $\log_{2}(x) = \log_{2}(8)$ so that $x = 8$ .

A second approach to solving $\log_{2}(x) = 3$ us to apply the corresponding exponential function, $f(x) = 2^x$ to both sides: $2^{\log_{2}(x)} = 2^{3}$ so $x = 2^3 = 8$ .

A third approach to solving $\log_{2}(x) = 3$ is to use Theorem 5.6 to rewrite $\log_{2}(x) = 3$ as $2^{3} = x$ , so $x=8$ .

In the grand scheme of things, all three approaches we have presented to solve $\log_{2}(x) = 3$ are mathematically equivalent, so we opt to choose the last approach in our summary below.

Steps for Solving an Equation Involving Logarithmic Functions

Isolate the logarithmic function.
(a) If convenient, express both sides as logs with the same base and equate
arguments.
(b) Otherwise, rewrite the log equation as an exponential equation.

Example 5.6.1

Example 5.6.1.1

Solve the following equations. Check your solutions graphically.

$\log_{117}(1-3x) = \log_{117}\left(x^2-3\right)$

Solution:

Solve $\log_{117}(1-3x) = \log_{117}\left(x^2-3\right)$ for $x$ .

We have the same base on both sides of the equation $\log_{117}(1-3x) = \log_{117}\left(x^2-3\right)$ , thus we equate the arguments (what’s inside) of the logs to get

$\begin{array}{rcl} 1-3x & = & x^2-3 \\ x^2+3x-4 & = & 0 \\ x=-4 & \text{ and } & x=1 \\ \end{array}$

To check these answers using a graph, we make use of the change of base formula and graph $\textcolor{red}{f(x) = \frac{\ln(1-3x)}{\ln(117)}}$ and $\textcolor{blue}{g(x) = \frac{\ln\left(x^2-3\right)}{\ln(117)}}$ . We see these graphs intersect only at $x=-4$ . however.

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To see what happened to the solution $x=1$ , 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 $x=1$ is not in the domain of the original equation, and is not a solution.

Example 5.6.1.2

Solve the following equations. Check your solutions graphically.

$2 - \ln(t-3) = 1$

Solution:

Solve $2 - \ln(t-3) = 1$ for $t$ .

To solve $2 - \ln(t-3) = 1$ , we first isolate the logarithm and get $\ln(t-3) = 1$ . Rewriting $\ln(t-3) = 1$ as an exponential equation, we get is $e^{1} = t-3$ , so $t =e+3$ .

$\begin{array}{rclr} 2 - \ln(t-3) & = & 1 & \text{Isolate the logarithm} \\ \ln(t-3) & = & 1 & \\[4pt] e^{1} & = & t-3 & \text{Rewrite as an exponential} \\ t & = & e+3 & \end{array}$

A graph shows the graphs of $\textcolor{red}{f(t) = 2 - \ln(t-3)}$ and $\textcolor{blue}{g(t) = 1}$ intersect at $t = e+3 \approx 5.718$ .

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Example 5.6.1.3

Solve the following equations. Check your solutions graphically.

$\log_{6}(x+4) + \log_{6}(3-x) = 1$

Solution:

Solve $\log_{6}(x+4) + \log_{6}(3-x) = 1$ for $x$ .

We start solving $\log_{6}(x+4) + \log_{6}(3-x) = 1$ by using the Product Rule for logarithms to rewrite the equation

$\begin{array}{rclr} \log_{6}\left[(x+4)(3-x)\right] &= & 1 & \text{Rewrite as an exponential} \\ 6^{1} & = & (x+4)(3-x) & \\ x^2+x-6 &=& 0 & \end{array}$

We get two solutions: $x=-3$ and $x=2$ .

Using the change of base formula, we graph $\textcolor{red}{y=f(x) = \frac{\ln(x+4)}{\ln(6)} + \frac{\ln(3-x)}{\ln(6)}}$ and $\textcolor{blue}{y=g(x) = 1}$ and we see the graphs intersect twice, at $x=-3$ and $x=2$ , as required.

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Example 5.6.1.4

Solve the following equations. Check your solutions graphically.

$\log_{7}(1-2t) = 1 - \log_{7}(3-t)$

Solution:

Solve $\log_{7}(1-2t) = 1 - \log_{7}(3-t)$ for $t$ .

Taking a cue from the previous problem, we begin solving $\log_{7}(1-2t) = 1 - \log_{7}(3-t)$ by first collecting the logarithms on the same side,

$\begin{array}{rclr} \log_{7}(1-2t) + \log_{7}(3-t) & = & 1 & \text{Power Rule} \\ \log_{7}[(1-2t)(3-t)] & = & 1 & \text{Rewrite as an exponential} \\[2pt] 7^{1} & =& (1-2t)(3-t) & \\ 7 & = & 2t^2 - 7t + 3 & \\ 2t^2-7t-4 &= &0 & \\ \end{array}$

Solving, we find $t = -\frac{1}{2}$ and $t=4$ .

Once again, we use the change of base formula and find the graphs of $\textcolor{red}{y = f(t) = \frac{\ln(1-2t)}{\ln(7)}}$ and $\textcolor{blue}{y=g(t) = 1 - \frac{\ln(3-t)}{\ln(7)}}$ intersect only at $t=-\frac{1}{2}$ .

Checking $t=4$ in the original equation produces $\log_{7}(-7) = 1 - \log_{7}(-1)$ , showing $t=4$ is not in the domain of $f$ nor $g$ .

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Example 5.6.1.5

Solve the following equations. Check your solutions graphically.

$\log_{2}(x+3) = \log_{2}(6-x)+3$

Solution:

Solve $\log_{2}(x+3) = \log_{2}(6-x)+3$ for $x$ .

Our first step in solving $\log_{2}(x+3) = \log_{2}(6-x)+3$ is to gather the logarithms to one side of the equation:

$\begin{array}{rclr} \log_{2}(x+3) - \log_{2}(6-x) & = & 3 & \\ \log_{2}\left(\frac{x+3}{6-x}\right) & = & 3 & \text{Quotient Rule} \\[2pt] 2^{3} & = & \frac{x+3}{6-x} & \text{Rewrite as an exponential} \\ 8(6-x) & = & x+3 & \\ x & = & 5 & \\ \end{array}$

Using the change of base once again, we graph $\textcolor{red}{f(x) = \frac{\ln(x+3)}{\ln(2)}}$ and $\textcolor{blue}{g(x) = \frac{\ln(6-x)}{\ln(2)} + 3}$ and find they intersect at $x=5$ .

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Example 5.6.1.6

Solve the following equations. Check your solutions graphically.

$1 + 2 \log_{4}(t+1) = 2 \log_{2}(t)$

Solution:

Solve $1 + 2 \log_{4}(t+1) = 2 \log_{2}(t)$ for $t$ .

Our first step in solving $1 + 2 \log_{4}(t+1) = 2 \log_{2}(t)$ is to gather the logs on one side of the equation. We obtain $1 = 2 \log_{2}(t) - 2 \log_{4}(t+1)$ but find we need a common base to combine the logs.

As $4$ is a power of $2$ , we use change of base to convert $\log_{4}(t+1) = \frac{\log_{2}(t+1)}{\log_{2}(4)} = \frac{1}{2} \log_{2}(t+1)$ . Hence, our original equation becomes

$\begin{array}{rclr} 1 & = & 2 \log_{2}(t) - 2 \left(\frac{1}{2} \log_{2}(t+1)\right) & \\ [2pt] 1 &= & 2\log_{2}(t) - \log_{2}(t+1) & \\ [2pt] 1 & = & \log_{2}\left(t^2\right) - \log_{2}(t+1) & \text{Power Rule} \\ [6pt] 1 & = & \log_{2}\left( \dfrac{t^{2}}{t+1}\right) & \text{Quotient Rule} \\ \frac{t^{2}}{t+1} & = & 2 & \text{Rewrite as a exponential} \\ t^2 -2t-2 & = & 0 & \\ \end{array}$

Using the quadratic formula, we obtain $t = 1 \pm \sqrt{3}$ .

One last time, we use the change of base formula and graph $\textcolor{red}{f(t) = 1 + \frac{2\ln(t+1)}{\ln(4)}}$ and $\textcolor{blue}{g(t) = \frac{2 \ln(t)}{\ln(2)}}$ . We see the graphs intersect only at $t = 1 + \sqrt{3} \approx 2.732$ .

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Note the solution $t = 1 - \sqrt{3} < 0$ . Hence if substituted into the original equation, the term $2 \log_{2}\left(1 - \sqrt{3}\right)$ is undefined, which explains why the graphs intersect only once.

If nothing else, Example 5.6.1 demonstrates the importance of checking for extraneous solutions^[2] when solving equations involving logarithms. Even though we checked our answers graphically, extraneous solutions are easy to spot: any supposed solution which causes the argument of a logarithm to be negative must be discarded.

While identifying extraneous solutions is important, it is equally important to understand which machinations create the opportunity for extraneous solutions to appear. In the case of Example 5.6.1, extraneous solutions, by and large, result from using the Power, Product, or Quotient Rules. We encourage the reader to take the time to track each extraneous solution found in Example 5.6.1 backwards through the solution process to see at precisely which step it fails to be a solution.

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

Example 5.6.2

Example 5.6.2.1

The function $f(x) = \dfrac{\log(x)}{1-\log(x)}$ is one-to-one.

Write a formula for $f^{-1}(x)$ and check your answer graphically.

Solution:

Write a formula for $f^{-1}(x)$ and check your answer graphically.

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

$\begin{array}{rclr} y & = & f(x) & \\ y & = & \dfrac{\log(x)}{1-\log(x)} & \\[8pt] x & = & \dfrac{\log(y)}{1-\log(y)} & \text{Interchange $x$ and $y$.}\\[8pt] x\left(1-\log(y)\right) & = & \log(y) & \\ x - x\log(y) & = & \log(y) & \\ x & = & x \log(y) + \log(y) & \\ x & = & (x+1) \log(y) & \\ \dfrac{x}{x+1} & = & \log(y) & \\ y & = & 10^{\frac{x}{x+1}} & \text{Rewrite as an exponential equation.}\\ \end{array}$

We have $f^{-1}(x) = 10^{\frac{x}{x+1}}$ . Graphing $\textcolor{red}{f}$ and $\textcolor{blue}{f^{-1}}$ on the same graph produces the required symmetry about $y=x$ .

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Example 5.6.2.2

The function $f(x) = \dfrac{\log(x)}{1-\log(x)}$ is one-to-one.

Solve $\dfrac{\log(x)}{1-\log(x)} = 1$

Solution:

Solve $\dfrac{\log(x)}{1-\log(x)} = 1$ .

Recognizing $\frac{\log(x)}{1-\log(x)} = 1$ as $f(x) = 1$ , we have

$\begin{array}{rcl} x & = & f^{-1}(1) \\ & = & 10^{\frac{1}{1+1}} \\ & = & 10^{\frac{1}{2}} \\ & = & \sqrt{10} \end{array}$

To check our answer algebraically, first recall $\log(\sqrt{10}) = \log_{10}(\sqrt{10})$ . Next, we know $\sqrt{10} = 10^{\frac{1}{2}}$ . Hence, $\log_{10} \left(10^{\frac{1}{2}} \right) = \frac{1}{2} = 0.5$ . It follows that $\frac{\log(\sqrt{10})}{1-\log(\sqrt{10})} = \frac{0.5}{1-0.5} = \frac{0.5}{0.5} = 1$ , as required.

5.6.1 Section Exercises

In Exercises 1 – 24, solve the equation analytically.

$\log(3x-1) = \log(4-x)$
$\log_{2}\left(x^{3}\right) = \log_{2}(x)$
$\ln\left(8-t^2\right)=\ln(2-t)$
$\log_{5}\left(18-t^2\right) = \log_{5}(6-t)$
$\log_{3}(7-2x) = 2$
$\log_{\frac{1}{2}} (2x-1) = -3$
$\ln\left(t^2-99\right) = 0$
$\log(t^2-3t) = 1$
$\log_{125} \left(\dfrac{3x-2}{2x+3}\right)=\dfrac{1}{3}$
$\log\left(\dfrac{x}{10^{-3}}\right) = 4.7$
$-\log(x) = 5.4$
$10\log\left(\dfrac{x}{10^{-12}}\right) = 150$
$6-3\log_{5}(2t)=0$
$3\ln(t)-2=1-\ln(t)$
$\log_{3}(t - 4) + \log_{3}(t + 4) = 2$
$\log_{5}(2t + 1) + \log_{5}(t + 2) = 1$
$\log_{169}(3x + 7) - \log_{169}(5x - 9) = \dfrac{1}{2}$
$\ln(x+1) - \ln(x) = 3$
$2\log_{7}(t) = \log_{7}(2) + \log_{7}(t+12)$
$\log(t) - \log(2) = \log(t+8) - \log(t+2)$
$\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$
$\ln(\ln(x)) = 3$
$\left(\log(t)\right)^2=2\log(t)+15$
$\ln(t^{2}) = (\ln(t))^{2}$

In Exercises 25 – 30, state the domain of the function.

$r(x) = \dfrac{x}{1 - \ln(x)}$
$R(x) = \dfrac{x \ln(x)}{1 - \ln(x)}$
$s(t) = \sqrt{2 - \log(t)}$
$c(t) = (2 \ln(t) -1)^{\frac{2}{3}}$
$\ell(t) = \ln( \ln(t))$
$L(x) = \log\left( \dfrac{x \ln(x)}{1 - \ln(x)} \right)$
Solve $\ln(3 - y) - \ln(y) = 2x + \ln(5)$ for $y$ .
In Example 5.6.2 we found the inverse of to be .
1. Algebraically check our answer by verifying $\left(f^{-1} \circ f\right)(x) = x$ for all $x$ in the domain of $f$ and that $\left(f \circ f^{-1}\right)(x) = x$ for all $x$ in the domain of $f^{-1}$
2. Find the range of $f$ by finding the domain of $f^{-1}$ .
3. Let $g(x) = \dfrac{x}{1 - x}$ and $h(x) = \log(x)$ . Show that $f = g \circ h$ and $(g \circ h)^{-1} = h^{-1} \circ g^{-1}$ .
Let $f(x) = \dfrac{1}{2}\ln\left(\dfrac{1 + x}{1 - x}\right)$ . Compute $f^{-1}(x)$ and find its domain and range.

Section 5.6 Exercise Answers can be found in the Appendix … Coming soon

They do, however, represent the same family of complex numbers. We refer the reader to a course in Complex Variables. ↵
Recall that an extraneous solution is an answer obtained analytically which does not satisfy the original equation. ↵

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Functions, Trigonometry, and Systems of Equations Copyright © by Vanessa Coffelt is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Steps for Solving an Equation Involving Logarithmic Functions

Example 5.6.1

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Example 5.6.2

Solution:

Solution:

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