Question

Log^5(1/25)=-2 in exponential form

Answer 1

We'll use the follwowing property, which comes from the definition of a logarithm:

`\log _ab=c\Leftrightarrow a^c=b`

i.e, The logarithm with base a of b is c if, and only if a to the c power equals b.

Using this to translate

`\log _5((1)/(25))=-2`

Into exponential form, will yield:

`5^(-2)`

Because:

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

ANSWER:

`5^(-2)`