# How do you solve 9^(x+1) = 27?

Apr 12, 2016

$x = \frac{1}{2}$

#### Explanation:

We will use the following properties:

• $\log \left({a}^{x}\right) = x \log \left(a\right)$

• ${\log}_{a} \left({a}^{x}\right) = x$

${9}^{x + 1} = 27$

$\implies {\log}_{3} \left({9}^{x + 1}\right) = {\log}_{3} \left(27\right)$

$\implies \left(x + 1\right) {\log}_{3} \left(9\right) = {\log}_{3} \left(27\right)$

$\implies \left(x + 1\right) {\log}_{3} \left({3}^{2}\right) = {\log}_{3} \left({3}^{3}\right)$

$\implies 2 \left(x + 1\right) = 3$

$\implies x + 1 = \frac{3}{2}$

$\therefore x = \frac{1}{2}$

Apr 12, 2016

Alternative solution:

#### Explanation:

You can write everything in the same base:

${\left({3}^{2}\right)}^{x + 1} = {3}^{3}$

${3}^{2 x + 2} = {3}^{3}$

${3}^{2 x} = {3}^{3 - 2}$

${3}^{2 x} = {3}^{1}$

$2 x = 1$

$x = \frac{1}{2}$

Hopefully this helps!