# How do you solve log_[2](12b-21) - log_[2](b^2-3) = 2?

Jun 17, 2016

Use the rule ${\log}_{a} \left(n\right) - {\log}_{a} \left(m\right) = {\log}_{a} \left(\frac{n}{m}\right)$

#### Explanation:

${\log}_{2} \left(\frac{12 b - 21}{{b}^{2} - 3}\right) = 2$

$\frac{12 b - 21}{{b}^{2} - 3} = 4$

$12 b - 21 = 4 \left({b}^{2} - 3\right)$

$0 = 4 {b}^{2} - 12 b - 12 + 21$

$0 = 4 {b}^{2} - 12 b + 9$

$0 = 4 {b}^{2} - 6 b - 6 b + 9$

$0 = 2 b \left(2 b - 3\right) - 3 \left(2 b - 3\right)$

$0 = \left(2 b - 3\right) \left(2 b - 3\right)$

$b = \frac{3}{2}$

However, checking in the original equation, we find this solution doesn't work.

Therefore, the solution set is $\left\{x = \emptyset\right\}$.

Hopefully this helps!