Question

How do you simplify `5^(log_5x)`?

Answer 1

See below.

Explanation:

Make `y = 5^(log_5x)`

Apply `log_5` to both sides

`log_5y=log_5xlog_5 5=log_5x cdot 1=log_5 x`

Now `log_5y=log_5x->y=x` so

`5^(log_5x)=x`