Question

How do you solve `log_(1/7)x=-1`?

Answer 1

I tried this:

Explanation:

We can use the definition of log:

`log_ba=x`

so that:

`a=b^x`

In our case:
`log_(1/7)x=-1`
so that:
`x=(1/7)^-1=7`
using a property of exponents.