# How do you solve log_4(x+1) - log_4(x-2) = 3?

Dec 6, 2015

I found: $x = 2.04762$

#### Explanation:

You can use the property of logs that tells us:
$\log x - \log y = \log \left(\frac{x}{y}\right)$
to get:
${\log}_{4} \left(\frac{x + 1}{x - 2}\right) = 3$
Now we use the definition of log to change it into an exponential:
$\frac{x + 1}{x - 2} = {4}^{3}$
$\frac{x + 1}{x - 2} = 64$
$\left(x + 1\right) = 64 \left(x - 2\right)$
$x + 1 = 64 x - 128$
$63 x = 129$
$x = \frac{129}{63} = 2.04762$