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

Oct 12, 2016

$x = 2$

#### Explanation:

Start by applying the rule $a \log n = \log {n}^{a}$.

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

Now apply the difference rules of logarithms: ${\log}_{a} \left(n\right) - {\log}_{a} \left(m\right) = {\log}_{a} \left(\frac{n}{m}\right)$.

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

Convert to exponential form using the rule ${\log}_{a} \left(n\right) = b \to {a}^{b} = n$

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

${x}^{2} / \left(x - 1\right) = 4$

${x}^{2} = 4 \left(x - 1\right)$

${x}^{2} = 4 x - 4$

${x}^{2} - 4 x + 4 = 0$

$\left(x - 2\right) \left(x - 2\right) = 0$

$x = 2$

Checking in the original equation:

log_4(2^2) - log_4(2 - 1) =^? 1

log_4(4) - log_4(1) =^?1

$1 - 0 = 1$

The solution works!

Hopefully this helps!