How do you solve log_3(x+4)-log_3x=2?

Dec 14, 2015

$x = \frac{1}{2}$

Explanation:

The relevant law of logarithms that will help us here is the following:

${\log}_{z} A - {\log}_{z} B = {\log}_{z} \left(\frac{A}{B}\right)$

where $A$ and $B$ can be any expression and $z$ is the base.

Applying this law to our equation will give us

${\log}_{3} \left(\frac{x + 4}{x}\right) = 2$

From here we can simply raise $3$ to both sides of the equation.

${3}^{{\log}_{3} \left(\frac{x + 4}{x}\right)} = {3}^{2}$

On the left-hand side, the $3$ will cancel with the ${\log}_{3}$ leaving us with

$\frac{x + 4}{x} = 9$

Solving for $x$ yields $x = \frac{1}{2}$.