# How do you solve 27^(2x+1) = 9^(4x)?

Jun 10, 2016

$x = \frac{3}{2}$

#### Explanation:

$27 = {3}^{3}$ and $9 = {3}^{2}$

then

${27}^{2 x + 1} = {9}^{4 x} \equiv {\left({3}^{3}\right)}^{2 x + 1} = {\left({3}^{2}\right)}^{4 x}$

or

${3}^{3 \left(2 x + 1\right)} = {3}^{8 x}$

so

$3 \left(2 x + 1\right) = 8 x$ Solving for $x$ gives $x = \frac{3}{2}$