# How do you solve 16^(3x) = 64^x + 9?

May 19, 2016

$x = 0.3041$, nearly.

#### Explanation:

The equation can be reorganized as ${\left({4}^{3 x}\right)}^{2} - {4}^{3 x} - 9 = 0$.

This is a quadratic in ${4}^{3 x}$. Solving,

${4}^{3 x} = {\left({4}^{3}\right)}^{x} = {64}^{x} = \frac{1 \pm \sqrt{37}}{2} = 3.5414 , - 2.5414$, nearly.

As #64^x>0, the negative root is inadmissible.

So, inversely, $x = \log \frac{3.5414}{\log} 64 = 0.3041$, nearly. .