Question

How do you solve `3^(x²+4x) = 1/81`?

Answer 1

I found: `x=-2`

Explanation:

Consider that `1/81=3^-4`
so you can write:
`3^(x^2+4x)=3^-4`
taking the log in base `3` on both sides you get (remembering that `log_a(a^x)=x)`:
`x^2+4x=-4`
and:
`x^2+4x+4=0`
Using the Quadratic Formula:
`x_(1,2)=(-4+-sqrt(16-16))/2=-4/2=-2`