Question

How do you solve for x in `3/2log_7x=log_7 125`?

Answer 1

I found `x=25`

Explanation:

Let us use a property of logs to get rid of the `3/2` as:
`log_7(x)^(3/2)=log_7(125)`
the logs are equal if the arguments are equal. So we need that:
`x^(3/2)=125`
let us take the exponent of `2/3` on both sides:
`x^(3/2*2/3)=125^(2/3)`
`x=125^(2/3)`
`x=root3(125^2)=25`