# How do you solve log_2 x + log_2 (x + 2) = 3?

Jan 27, 2016

Use properties of logarithms and exponentiation to derive a quadratic equation, one of whose roots is a solution to the original equation, namely $x = 2$

#### Explanation:

If $a , b , c > 0$ then ${\log}_{c} \left(a\right) + {\log}_{c} \left(b\right) = {\log}_{c} \left(a b\right)$

Hence if $x > 0$ then ${\log}_{2} \left(x\right) + {\log}_{2} \left(x + 2\right) = {\log}_{2} \left(x \left(x + 2\right)\right)$

So we find:

$8 = {2}^{3} = {2}^{{\log}_{2} \left(x\right) + {\log}_{2} \left(x + 2\right)} = {2}^{{\log}_{2} \left(x \left(x + 2\right)\right)} = x \left(x + 2\right)$

Rearranging slightly, this becomes:

$0 = {x}^{2} + 2 x - 8 = \left(x + 4\right) \left(x - 2\right)$

So $x = - 4$ or $x = 2$

$x = 2$ is a solution of the original equation, as we can check if we wish:

${\log}_{2} 2 + {\log}_{2} \left(2 + 2\right) = 1 + 2 = 3$

$x = - 4$ does not satisfy $x > 0$ so we are not guaranteed that it will work, even if we allow Complex logarithms. In fact we find:

${\log}_{2} \left(- 4\right) + {\log}_{2} \left(- 4 + 2\right)$

$= \left(2 + \frac{\pi}{\ln} \left(2\right) i\right) + \left(1 + \frac{\pi}{\ln} \left(2\right) i\right) = 3 + \frac{2 \pi}{\ln} \left(2\right) i \ne 3$

So the only solution of the original equation is $x = 2$