Question

How do you solve `25^ { x } = 5^ { x ^ { 2} - 3}`?

Answer 1

`x = 3, x = -1`

Explanation:

`25` is the same as `5^2`, so

`25^x = 5^(2x)`.

Now since the bases are equal, we can solve for the exponents:

`2x = x^2 - 3`
`x^2 - 2x - 3 = 0`
`(x - 3)(x + 1) = 0`
`x = 3, x = -1`