Question

How do you solve `5^x = 71`?

Answer 1

`x=log_5(71)`

Explanation:

We will be using the following properties of logarithms:

• `log(a^x) = xlog(a)`

• `log_a(a) = 1`

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

`5^x = 71`

`=>log_5(5^x) = log_5(71)`

`=>xlog_5(5)=log_5(71)`

`=>x*1=log_5(71)`

`=>x=log_5(71)`