Question

How do you write `log_(1/25)5=-1/2` in exponential form?

Answer 1

`(1/25)^(-1/2)=5`

Explanation:

`log_ab=x` in exponential form is written as `a^x=b`

Hence `log_(1/25)5=-1/2` can be written in exponential form as `(1/25)^(-1/2)=5`

Answer 2

`5=(1/25)^(-1/2)`

Explanation:

`"using the "color(blue)"law of logarithms"`

`•color(white)(x)log_b x=nhArrx=b^n`

`"here "x=5,b=1/25" and "n=-1/2`

`rArrlog_(1/25)5=-1/2hArr5=(1/25)^(-1/2)`