Question

How do you evaluate `log_225 15`?

Answer 1

`1/2`.

Explanation:

`x=log_(225) 15 rArr 225^x=15 rArr (15^2)^x=15rArr 15^(2x)=15^1`

#rArr 2x=1

`rArr x=1/2`

`rArr log_(225) 15=1/2`.