# How do you solve 2^(2x)times4^(4x+8)=64?

Sep 18, 2016

$x = - 1$

#### Explanation:

We have: ${2}^{2 x} \times {4}^{4 x + 8} = 64$

Let's express the numbers in terms of $2$:

$\implies {2}^{2 x} \times {\left({2}^{2}\right)}^{4 x + 8} = {2}^{6}$

Using the laws of exponents:

$\implies {2}^{2 x} \times {2}^{8 x + 16} = {2}^{6}$

$\implies {2}^{2 x + 8 x + 16} = {2}^{6}$

$\implies {2}^{10 x + 16} = {2}^{6}$

$\implies 10 x + 16 = 6$

$\implies 10 x = - 10$

$\implies x = - 1$

Therefore, the solution to the equation is $x = - 1$.