# How do you solve log_10a+log_10(a+21)=2?

Nov 5, 2016

$a = 4$.

#### Explanation:

Use the rule ${\log}_{a} \left(n\right) + {\log}_{a} \left(m\right) = {\log}_{a} \left(n \times m\right)$.

${\log}_{10} \left(a \left(a + 21\right)\right) = 2$

${a}^{2} + 21 a = 100$

${a}^{2} + 21 a - 100 = 0$

$\left(a + 25\right) \left(a - 4\right) = 0$

$a = - 25 \mathmr{and} a = 4$

However, $a = - 25$ is extraneous, since it renders the original equation undefined. The only actual solution is $a = 4$.