# How do you solve 3^(x²+4x) = 1/81?

Jul 16, 2015

I found: $x = - 2$

#### Explanation:

Consider that $\frac{1}{81} = {3}^{-} 4$
so you can write:
${3}^{{x}^{2} + 4 x} = {3}^{-} 4$
taking the log in base $3$ on both sides you get (remembering that log_a(a^x)=x):
${x}^{2} + 4 x = - 4$
and:
${x}^{2} + 4 x + 4 = 0$
${x}_{1 , 2} = \frac{- 4 \pm \sqrt{16 - 16}}{2} = - \frac{4}{2} = - 2$