# How do you solve log3x=log2+log (x+5)?

Apr 8, 2016

$x = 10$

#### Explanation:

Rearrange slightly to get $x$ on one side and constant on the other

$\log 3 x = \log 2 + \log \left(x + 5\right)$
$\log 3 + \log x = \log 2 + \log \left(x + 5\right)$
$\log x - \log \left(x + 5\right) = \log 2 - \log 3$

Now, using laws of logarithms, make it so you only have a single logarithm on either side

$\log x - \log \left(x + 5\right) = \log \left(\frac{x}{x + 5}\right)$
$\log 2 - \log 3 = \log \left(\frac{2}{3}\right)$

So you have

$\log \left(\frac{x}{x + 5}\right) = \log \left(\frac{2}{3}\right)$

Raise both sides to the power $e$ to cancel out the logarithms (assuming we are dealing with ${\log}_{e}$ or $\ln$),

$\frac{x}{x + 5} = \frac{2}{3}$

From which we get

$\frac{x}{x + 5} = \frac{2}{3}$
$3 x = 2 \left(x + 5\right)$
$3 x = 2 x + 10$
$3 x - 2 x = 10$
$x = 10$