How do you solve log_2(6x+9)-log_2(3)=3?

Dec 24, 2015

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

Explanation:

Use the following rule: ${\log}_{a} \left(b\right) - {\log}_{a} \left(c\right) = {\log}_{a} \left(\frac{b}{c}\right)$

${\log}_{2} \left(6 x + 9\right) - {\log}_{2} \left(3\right) = 3$

$\implies {\log}_{2} \left(\frac{6 x + 9}{3}\right) = 3$

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

Exponentiate both sides with base $2$ to undo the logarithm.

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

$2 x + 3 = 8$

$2 x = 5$

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