Solving equations using logarithms

In a nutshell

Given an equation of the form $a^x = b$ , where $a$ is a positive constant not equal to $1$ , and $b$ is a positive constant, you can use logarithms to solve for $x$ .

Logarithms as inverses to exponentials

Recall that both $\log_a(a^x) = x$ , and that $a^{\log_a(x)} =x$ , i.e. the functions $a^x$ and $\log_a(x)$ are inverse to each other. Consequently, you can apply $\log_a()$ to both sides of an exponential equation to solve it for the variable in the exponent.

Given an exponential equation, e.g.:

$a^x = b$

Apply the logarithm with base $a$ to both sides of the equation:

$\begin{aligned}a^x &= b\\\log_a(a^x) &= \log_a(b)\\x &= \log_a(b)\end{aligned}$

Yielding the desired solution.

Example 1

Solve the equation $5^{2x-1} = 10$ , giving your solution to $3$ decimal places.

Rearrange the equation by multiplying by $5$ and taking square roots:

$\begin{aligned}5^{2x-1} &= 10\\5^{2x} &= 50\\(5^x)^2 &= 50\\5^x &= \sqrt{50}\\\end{aligned}$ 

Apply $\log_5()$ to both sides of the equation:

$\begin{aligned}5^x &= \sqrt{50}\\\log_5(5^x) &= \log_5(\sqrt{50})\\x &= 1.2153\dots\end{aligned}$ 

Therefore, to $3$ decimal places, the solution to the equation $5^{2x-1} = 10$ is $\underline{1.215}$ .

Example 2

Find the exact values of all solutions to the equation $9^x = 3^{x+1} -2$ .

Observe that $9^x = (3^2)^x = (3^x)^2$ , and similarly that $3^{x+1} = 3 \times 3^x$ . Rearrange the equation to get a quadratic in $3^x$ :

$\begin{aligned}9^x &= 3^{x+1} - 2\\(3^x)^2 - 3 \times 3^x + 2 &= 0\\(3^x - 1)(3^x - 2) & = 0\end{aligned}$ 

Hence the solutions to this equation are the solutions to $3^x = 1$ and $3^x = 2$ . Applying $\log_3()$ to these, get that these solutions are $x = \log_3(1) = 0$ and $x = \log_3(2)$ .

Therefore, the exact values of the solutions to the equation $9^x = 3^{x+1} - 2$ are $\underline{x=0}$ and $\underline{x=\log_3(2)}$ .

Given any equation of the form $f(x) = g(x)$ , you can apply the logarithm with base $a$ to both sides of the equation and get the equation $\log_a(f(x)) = \log_a(g(x))$ , allowing you to solve more complicated equations.

Example 3

Find the solution to the equation $2^x = 5^{3x-2}$ . Give your answer to $3$ decimal places.

Apply $\ln()$ to both sides of the equation (you can apply a logarithm with any base, but $\ln()$ is the natural logarithm, so serves as a "natural" choice when a choice of base does not matter).

\begin{aligned}2^x &= 5^{3x-2}\\\ln(2^x) &= \ln(5^{3x-2})\\x\ln(2) &= (3x-2)\ln(5)\\(3\ln(5) - \ln(2))x &= 2\ln(5)\\x &= \dfrac{2\ln(5)}{3\ln(5) - \ln(2)}\\x &= 0.7784\dots\end{aligned}

Therefore, to $3$ decimal places, the solution to the equation $2^x = 5^{3x-2}$ is $\underline{x = 0.778}$ .