# How do you solve log _10 (x+21) +log_10 (x)= 2?

May 18, 2016

See below.

#### Explanation:

In general: ${\log}_{n} \left(a\right) + {\log}_{n} \left(b\right) = {\log}_{n} \left(a b\right)$
Therefore: ${\log}_{10} \left(x + 21\right) + {\log}_{10} \left(x\right) = {\log}_{10} \left({x}^{2} + 21 x\right)$
By the definition of a logarithm:
${\log}_{10} \left({x}^{2} + 21 x\right) = 2 \implies {10}^{2} = {x}^{2} + 21 x$
By solving this as a quadratic:
$100 = {x}^{2} + 21 x$
${x}^{2} + 21 x - 100 = 0$
$\left(x - 4\right) \left(x + 25\right) = 0$
$x = 4$ or $x = - 25$ (are solutions to the equation), but obviously you can't have a negative within a logarithm so x=-25 is not a solution.

It's always important to know the rules of what's being dealt with before attempting it.