Question

How do you solve `3^ { x } = \sqrt { 5^ { x - 2} }`?

Answer 1

`x=log_(5/9)(25)`

Explanation:

Let's rewrite the equation:
`3^x=(5^(x-2))^(1/2)`

`(3^x)^2=(5^(x-2))^(1/2*2)`
`3^(2x)=(5^x)/5^2`
`(5^x)/3^(2x)=25`
`(5/9)^x=25`

`x=log_(5/9)(25)~~-5.48`

If you are unfamiliar with the concept of logarithm, here's a quick overview:

If we have `b^x=y`
Then, `x=log_(b)(y)`, pronounced "log base b of y."