Question

How do you write `log_7 (1/49)=-2` in exponential form?

Answer 1

see explaination

Explanation:

Let say `log_7(1/49)=y`

`(1/49) = 7^y` use concept if `log_ab=c`, then `b=a^c`

`1/7^2=7^y`

`7^-2=7^y`

`y=-2`

therefore, `log_7(1/49)=-2`