# How do you solve log_4x- log_4(x-1)= 1/2?

Feb 27, 2016

$x = 2$

#### Explanation:

$1$. Recall the log quotient rule: ${\log}_{\textcolor{red}{b}} \left(\frac{\textcolor{b l u e}{m}}{\textcolor{p u r p \le}{n}}\right) = {\log}_{\textcolor{red}{b}} \left(\textcolor{b l u e}{m}\right) - {\log}_{\textcolor{red}{b}} \left(\textcolor{p u r p \le}{n}\right)$. Using the formula, rewrite the left side of the equation.

${\log}_{4} \left(x\right) - {\log}_{4} \left(x - 1\right) = \frac{1}{2}$

${\log}_{4} \left(\frac{x}{x - 1}\right) = \frac{1}{2}$

$2$. Rewrite the equation in exponential form.

${4}^{\frac{1}{2}} = \frac{x}{x - 1}$

$3$. Solve for $x$.

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

$x = 2 \left(x - 1\right)$

$x = 2 x - 2$

$\textcolor{g r e e n}{x = 2}$