# How do you solve (5^x)^(x+1)=5^(2x+12)?

Mar 9, 2016

$x = - 3$ or $x = 4$

#### Explanation:

As ${\left({a}^{m}\right)}^{n} = {a}^{m n}$, ${\left({5}^{x}\right)}^{x + 1} = {5}^{x \left(x + 1\right)} = {5}^{{x}^{2} + x}$

Hence, ${5}^{{x}^{2} + x} = {5}^{2 x + 12}$ or

${x}^{2} + x = 2 x + 12$ or ${x}^{2} + x - 2 x - 12 = 0$ or

${x}^{2} - x - 12 = 0$ or

${x}^{2} - 4 x + 3 x - 12 = 0$ or

$x \left(x - 4\right) + 3 \left(x - 4\right) = 0$ or

$\left(x + 3\right) \left(x - 4\right) = 0$ or

$x = - 3$ or $x = 4$