# How do you solve 3^(2x+2) + 3^(x+1) - 12 = 0?

Aug 16, 2015

Recognise as a quadratic in ${3}^{x + 1}$, one of whose solutions gives us a Real solution, $x = 0$.

#### Explanation:

Let $t = {3}^{x + 1}$

Then $0 = {3}^{2 x + 2} + {3}^{x + 1} - 12 = {t}^{2} + t - 12 = \left(t + 4\right) \left(t - 3\right)$

So $t = - 4$ or $t = 3$.

Now ${3}^{x + 1} > 0$ for all $x \in \mathbb{R}$, so we can discard the case $t = - 4$.

The remaining solution gives us:

${3}^{x + 1} = t = 3 = {3}^{1}$

Since exponentiation is one-one as a function from $\mathbb{R}$ to $\left(0 , \infty\right)$, this implies $x + 1 = 1$, so $x = 0$.