Question

How do you solve `7^x = 80`?

Answer 1

well, by inspection we know that `7^2=49 and 7^3=343`
so this means that the exponent 'x' must be between 2 and 3 (and closer to 2 than to 3).

so we convert from exponent form to log form and we obtain: `log_7(80)=x` which can be solved on a calculator or by using the change of base rule: `log80/log7`
or approximately 2.25