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).​

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