# How do you solve 2log_8(2x-5)=4?

Nov 25, 2015

$x = \frac{69}{2}$

#### Explanation:

First of all, your domain is $2 x - 5 > 0 \iff x > \frac{5}{2}$ since the argument of any logarithmic expression needs to be greater than zero.

Now, to solve the equation, you should first divide both sides of the equation by $2$:

$\textcolor{w h i t e}{\times} 2 {\log}_{8} \left(2 x - 5\right) = 4$
$\iff {\log}_{8} \left(2 x - 5\right) = 2$

The inverse function of ${\log}_{8} \left(x\right)$ is ${8}^{x}$. This means that ${\log}_{8} \left({8}^{x}\right) = x$ and ${8}^{{\log}_{8} x} = x$.

In other words, you can make ${\log}_{8}$ disappear by applying the exponential function ${8}^{x}$ on both sides!

$\iff {8}^{{\log}_{8} \left(2 x - 5\right)} = {8}^{2}$

$\iff 2 x - 5 = 64$

$\iff 2 x = 69$

$\iff x = \frac{69}{2}$