# How do you solve 2log _2 x + log_2 5=log_2 125?

Dec 4, 2015

Apply properties of logarithms to find that $x = 5$

#### Explanation:

We use the following properties of logarithms:
${\log}_{2} \left({a}^{b}\right) = b {\log}_{2} \left(a\right) \text{ }$ (for $a > 0$)

${\log}_{2} \left(\frac{a}{b}\right) = {\log}_{2} \left(a\right) - {\log}_{2} \left(b\right) \text{ }$ (for $a , b > 0$)

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

Before we begin, note that as we start by taking a logarithm of $x$, we know that $x > 0$.

$2 {\log}_{2} \left(x\right) + {\log}_{2} \left(5\right) = {\log}_{2} \left(125\right)$

$\implies {\log}_{2} \left({x}^{2}\right) + {\log}_{2} \left(5\right) = {\log}_{2} \left(125\right)$

$\implies {\log}_{2} \left({x}^{2}\right) = {\log}_{2} \left(125\right) - {\log}_{2} \left(5\right)$

$\implies {\log}_{2} \left({x}^{2}\right) = {\log}_{2} \left(\frac{125}{5}\right) = {\log}_{2} \left(25\right)$

$\implies {2}^{{\log}_{2} \left({x}^{2}\right)} = {2}^{{\log}_{2} \left(25\right)}$

$\implies {x}^{2} = 25$

$\implies x = \pm 5$

But we noted earlier that $x > 0$, so

$x = 5$