# What is the value of x if log_6 48 = log_6(x + 7) + log_6(x - 1)?

Mar 6, 2017

$\left\{5\right\}$

#### Explanation:

Combine the logarithms.

${\log}_{6} 48 = {\log}_{6} \left(\left(x + 7\right) \left(x - 1\right)\right)$

If ${\log}_{a} n = {\log}_{a} m$, then $n = m$.

$48 = \left(x + 7\right) \left(x - 1\right)$

$48 = {x}^{2} + 7 x - x - 7$

$48 = {x}^{2} + 6 x - 7$

$0 = {x}^{2} + 6 x - 55$

$0 = \left(x + 11\right) \left(x - 5\right)$

$x = - 11 \mathmr{and} 5$

$x = - 11$ is clearly extraneous as a logarithm can never be negative. $x = 5$ is valid, though.

Practice Exercises

1. Solve the following equations using ${\log}_{a} n - {\log}_{a} m = {\log}_{a} \left(\frac{n}{m}\right)$ and ${\log}_{a} m + {\log}_{a} n = {\log}_{a} \left(m \cdot n\right)$.

a) ${\log}_{2} \left(2 x\right) + {\log}_{2} \left(\frac{1}{2} x\right) = {\log}_{2} 64$
b) ${\log}_{3} \left(x + 2\right) - {\log}_{3} \left(x - 4\right) = 1$

a) $\left\{8\right\}$
b) $\left\{7\right\}$