Question

How do you solve for x in `(125)^x = 625`?

Answer 1

`x=4/3=1.333`

Explanation:

`125^x=625` means `x=log_(125)625`

As `log_ba=loga/logb`

`x=log625/log125=2.79588/2.09691=1.333`

Alternately - `125^x=625hArr(5^3)^x=5^4hArr5^(3x)=5^4` or

`3xlog5=4log5` or `x=4/3=1.333`