# How do you solve 2log_2x-log_2 5=2^2?

Apr 10, 2016

$x = 4 \sqrt{5}$

#### Explanation:

As ${\log}_{b} a = \log \frac{a}{\log} b$

$2 {\log}_{2} x - {\log}_{2} 5 = {2}^{2}$ can be written as

$2 \log \frac{x}{\log} 2 - \log \frac{5}{\log} 2 = 4$ and

multiplying each side by $\log 2$ as $\log 2 \ne 0$, we get

$2 \log x - \log 5 = 4 \log 2$ or

${x}^{2} / 5 = {2}^{4}$ or ${x}^{2} = 80$ or

$x = \sqrt{80} = 4 \sqrt{5}$ (as $x$ cannot be negative)

Apr 10, 2016

$\textcolor{b l u e}{\implies x = 4 \sqrt{5}}$

#### Explanation:

Example: I have chosen to use log to base 10 so that you can check it on your calculator if you so wish.

suppose we had $L o {g}_{10} \left(z\right) = 2$

This means$\to {10}^{2} = z$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the value of } x}$

Given: $\text{ } 2 {\log}_{2} x - {\log}_{2} 5 = {2}^{2}$

Write as: ${\log}_{2} \left({x}^{2}\right) - {\log}_{2} \left(5\right) = 4$

Subtraction of logs means that the source value are applying division

$\implies {\log}_{2} \left({x}^{2} / 5\right) = 4$

Using the principle demonstrated in my example

$\implies {\log}_{2} \left({x}^{2} / 5\right) = 4 \text{ "->" } {2}^{4} = {x}^{2} / 5$

$\implies x = \sqrt{80} \text{ "=" } \sqrt{{2}^{2} \times {2}^{2} \times 5}$

$\textcolor{b l u e}{\implies x = 4 \sqrt{5}}$