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# 14.2: Solving Exponential and Logarithmic Equations

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We can solve exponential and logarithmic equations by applying logarithms and exponentials. Since the exponential and logarithmic functions are invertible (they are inverses of each other), applying these operations can also be reversed. In short we have observed that

## Observation: Exponential and Logarithmic

$x=y \quad \Longrightarrow \quad b^{x}=b^{y}, \quad \text { and } \quad b^{x}=b^{y} \quad \Longrightarrow \quad x=y \nonumber$

$x=y \quad \Longrightarrow \quad \log _{b}(x)=\log _{b}(y), \quad \text { and } \quad \log _{b}(x)=\log _{b}(y) \quad \Longrightarrow \quad x=y \nonumber$

We can use the above implications whenever we have a common basis, or when we can convert the terms to a common basis.

## Example $$\PageIndex{1}$$

Solve for $$x$$.

1. $$2^{x+7}=32$$
2. $$10^{2x-8}=0.01$$
3. $$7^{2x-3}=7^{5x+4}$$
4. $$5^{3x+1}=25^{4x-7}$$
5. $$\ln(3x-5)=\ln(x-1)$$ & f)
6. $$\log_2(x+5)=\log_2(x+3)+4$$
7. $$\log_6(x)+\log_6(x+4)=\log_6(5)$$
8. $$\log_3(x-2)+\log_3(x+6)=2$$

Solution

In these examples, we can always write both sides of the equation as an exponential expression with the same base.

1. $$2^{x+7}=32 \quad \Longrightarrow \quad 2^{x+7}=2^{5} \Longrightarrow \quad x+7=5 \Longrightarrow \quad x=-2$$
2. $$10^{2 x-8}=0.01 \Longrightarrow 10^{2 x-8}=10^{-2} \Longrightarrow \quad 2 x-8=-2 \Longrightarrow 2 x=6 \Longrightarrow \quad x=3$$

Here it is useful to recall the powers of $$10$$, which were also used to solve the equation above.

$\boxed{ \begin{matrix} 10^3 &=&1,000 \\ 10^2&=&100 \\ 10^1&=&10 \\ 10^0&=&1 \\ 10^{-1}&=&0.1 \\ 10^{-2}&=&0.01 \\ 10^{-3}&=&0.001 \end{matrix} \quad\quad \text{In general } (n\geq1):\quad\left\{ \begin{matrix} 10^n=1\underbrace{00\cdots00}_{n\text{ zeros}} \\ \quad \\ 10^{-n}=\underbrace{0.0\cdots 00}_{n\text{ zeros}}1 \end{matrix}\right.} \nonumber$

1. $$7^{2 x-3}=7^{5 x+4} \Longrightarrow 2 x-3=5 x+4 \stackrel{(-5 x+3)}{\Longrightarrow} \quad-3 x=7 \Longrightarrow \quad x=-\dfrac{7}{3}$$
2. $$5^{3 x+1}=25^{4 x-7} \Longrightarrow 5^{3 x+1}=5^{2 \cdot(4 x-7)} \Longrightarrow 3 x+1=2 \cdot(4 x-7) \Longrightarrow 3 x+1=8 x-14 \stackrel{(-8 x-1)}{\Longrightarrow} \quad-5 x=-15 \Longrightarrow x=3$$

By a similar reasoning we can solve equations involving logarithms whenever the bases coincide.

1. $$\ln (3 x-5)=\ln (x-1) \Longrightarrow 3 x-5=x-1 \stackrel{(-x+5)}{\Longrightarrow} \quad 2 x=4 \Longrightarrow x=2$$
2. For part (f), we have to solve $$\log_2(x+5)=\log_2(x+3)+4$$. To combine the right-hand side, recall that $$4$$ can be written as a logarithm, $$4=\log_2(2^4)=\log_2 16$$. With this remark we can now solve the equation for $$x$$.

\begin{aligned} \log_2(x+5) &= \log_2(x+3)+4\\ \implies \quad \log_2(x+5)&=\log_2(x+3)+\log_2(16) \\ \implies \quad \log_2(x+5)&=\log_2(16\cdot (x+3)) \\ \implies \qquad \quad \;\; x+5&=16(x+3) \\ \implies \qquad \quad \;\; x+5&=16x+48 \\ \quad \stackrel{(-16x-5)}\implies \qquad \;\;-15x&=43\\ \implies \qquad \qquad \;\;x&=-\dfrac{43}{15} \end{aligned} \nonumber

1. Next, in part (g), we start by combining the logarithms. \begin{aligned} \log_6(x)+\log_6(x+4)&=\log_6(5) \\ \implies \quad \log_6(x(x+4))&=\log_6(5) \\ \stackrel{\text{remove }\log_6}\implies \qquad x(x+4)&=5 \\ \implies \qquad x^2+4x-5&=0 \\ \implies \quad(x+5)(x-1)&=0 \\ \qquad\implies x=-5 \text{ or } x=1\end{aligned} \nonumber

Since the equation became a quadratic equation, we ended up with two possible solutions $$x=-5$$ and $$x=1$$. However, since $$x=-5$$ would give a negative value inside a logarithm in our original equation $$\log_6(x)+\log_6(x+4)=\log_6(5)$$, we need to exclude this solution. The only solution is $$x=1$$.

1. Similarly we can solve the next part, using that $$2=\log_3(3^2)$$:

\begin{aligned} \log _{3}(x-2)+\log _{3}(x+6)&=2 \\ \Longrightarrow \log _{3}((x-2)(x+6))&=\log _{3}\left(3^{2}\right) \\ \Longrightarrow(x-2)(x+6)&=3^{2} \\ \Longrightarrow x^{2}+4 x-12&=9 \\ \Longrightarrow x^{2}+4 x-21&=0 \\ \Longrightarrow(x+7)(x-3)&=0 \\ \Longrightarrow x=-7 \text { or } x=3 \end{aligned} \nonumber

We exclude $$x=-7$$, since we would obtain a negative value inside a logarithm, so that the solution is $$x=3$$.

When the two sides of an equation are not exponentials with a common base, we can solve the equation by first applying a logarithm and then solving for $$x$$. Indeed, recall from Observation in section 7.2 on inverse functions, that since $$f(x)=\log_b(x)$$ and $$g(x)=b^x$$ are inverse functions, we have $$\log_b(b^x)=x$$ and $$b^{\log_b(x)}=x$$ whenever the compositions of the left sides make sense. That is, the action of the logarithm cancels out the action of the exponential function, and vice versa. So we can think of applying a logarithm (an exponentiation) on both sides of an equation to cancel an exponentiation (a logarithm) much like squaring both sides of an equation to cancel a square root.

## Example $$\PageIndex{2}$$

Solve for $$x$$.

1. $$3^{x+5}=8$$
2. $$13^{2x-4}=6$$
3. $$5^{x-7}=2^{x}$$
4. $$5.1^{x}=2.7^{2x+6}$$
5. $$17^{x-2}=3^{x+4}$$
6. $$7^{2x+3}=11^{3x-6}$$

Solution

We solve these equations by applying a logarithm (both $$\log$$ or $$\ln$$ will work for solving the equation), and then we use the identity $$\log(a^x)=x\cdot \log(a)$$.

1. $$3^{x+5}=8 \implies \ln 3^{x+5}=\ln 8 \implies (x+5)\cdot\ln 3=\ln 8 \implies x+5=\dfrac{\ln 8}{\ln 3}\implies x=\dfrac{\ln 8}{\ln 3}-5\approx -3.11$$
2. $$13^{2x-4}=6 \implies \ln 13^{2x-4}=\ln 6 \implies (2x-4)\cdot\ln 13=\ln 6 \implies 2x-4=\dfrac{\ln 6}{\ln 13} \implies 2x=\dfrac{\ln 6}{\ln 13}+4 \implies x=\dfrac{\frac{\ln 6}{\ln 13}+4}{2}=\dfrac{\ln 6}{2\cdot \ln 13}+2\approx 2.35$$
3. $$5^{x-7}=2^{x} \implies \ln 5^{x-7}=\ln 2^{x} \implies (x-7)\cdot\ln 5=x\cdot \ln 2$$

At this point, the calculation will proceed differently than the calculations in parts (a) and (b). Since $$x$$ appears on both sides of $$(x-7)\cdot\ln 5=x\cdot \ln 2$$, we need to separate terms involving $$x$$ from terms without $$x$$. That is, we need to distribute $$\ln 5$$ on the left and separate the terms. We have

\begin{aligned} (x-7)\cdot\ln 5=x\cdot \ln 2 & \implies & x\cdot\ln 5-7\cdot \ln 5=x\cdot \ln 2 \\ (\text{add }+7\cdot \ln 5-x\cdot \ln 2) &\implies & x\cdot\ln 5-x\cdot \ln 2=7\cdot \ln 5\\ &\implies & x\cdot(\ln 5-\ln 2)=7\cdot \ln 5 \\ &\implies & x=\dfrac{7\cdot \ln 5 }{\ln 5-\ln 2}\approx 12.30\end{aligned} \nonumber

We need to apply the same solution strategy for the remaining parts (d)-(f) as we did in (c).

1. \begin{aligned} 5.1^{x}=2.7^{2x+6} & \implies & \ln 5.1^{x}=\ln 2.7^{2x+6} \\ & \implies & x\cdot \ln 5.1=(2x+6)\cdot \ln 2.7 \\ & \implies & x\cdot \ln 5.1=2x\cdot \ln 2.7+6\cdot \ln 2.7 \\ & \implies & x\cdot \ln 5.1-2x\cdot \ln 2.7=6\cdot \ln 2.7 \\ & \implies & x\cdot (\ln 5.1-2\cdot \ln 2.7) = 6 \cdot \ln 2.7 \\ & \implies & x = \frac{6 \cdot \ln 2.7} {\ln 5.1-2\cdot \ln 2.7}\approx -16.68\end{aligned} \nonumber
1. \begin{aligned} 17^{x-2}=3^{x+4} & \implies & \ln 17^{x-2}=\ln 3^{x+4} \\ & \implies & (x-2)\cdot \ln 17=(x+4)\cdot \ln 3 \\ & \implies & x\cdot \ln 17-2\cdot \ln 17=x\cdot \ln 3+4\cdot \ln 3 \\ & \implies & x\cdot \ln 17-x\cdot \ln 3=2\cdot \ln 17+4\cdot \ln 3 \\ & \implies & x\cdot (\ln 17- \ln 3) =2\cdot \ln 17+4\cdot \ln 3 \\ & \implies & x = \dfrac{2\cdot \ln 17+4\cdot \ln 3} {\ln 17- \ln 3}\approx 5.80\end{aligned} \nonumber
1. \begin{aligned} 7^{2x+3}=11^{3x-6} & \implies & \ln 7^{2x+3}=\ln 11^{3x-6} \\ & \implies & (2x+3)\cdot \ln 7=(3x-6)\cdot \ln 11 \\ & \implies & 2x\cdot \ln 7+3\cdot \ln 7=3x\cdot \ln 11-6\cdot \ln 11 \\ & \implies & 2x\cdot \ln 7-3x\cdot \ln 11=-3\cdot \ln 7-6\cdot \ln 11 \\ & \implies & x\cdot (2\cdot \ln 7-3\cdot \ln 11)=-3\cdot \ln 7-6\cdot \ln 11 \\ & \implies & x = \dfrac{-3\cdot \ln 7-6\cdot \ln 11} {2\cdot \ln 7-3\cdot \ln 11}\approx 6.13\end{aligned} \nonumber

Note that in the problems above we could have also changed the base as we did earlier in the section. For example, in part f) above, we could have begun by writing the right hand side as $$11^{3x-6}=7^{\log_7 (11^{(3x-6)})}$$. We chose to simply apply a log to both sides instead, because the notation is somewhat simpler.

This page titled 14.2: Solving Exponential and Logarithmic Equations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Thomas Tradler and Holly Carley (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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