How do you solve 5^x = 71?

Feb 26, 2016

$x = {\log}_{5} \left(71\right)$

Explanation:

We will be using the following properties of logarithms:

• $\log \left({a}^{x}\right) = x \log \left(a\right)$

• ${\log}_{a} \left(a\right) = 1$

Then, to solve, we simply take the base-5 logarithm of each side of the equation.

${5}^{x} = 71$

$\implies {\log}_{5} \left({5}^{x}\right) = {\log}_{5} \left(71\right)$

$\implies x {\log}_{5} \left(5\right) = {\log}_{5} \left(71\right)$

$\implies x \cdot 1 = {\log}_{5} \left(71\right)$

$\implies x = {\log}_{5} \left(71\right)$