# How do you solve log_2(x/2)=5?

Jul 15, 2016

We raise both sides of our equation to be the exponent of a base of $2$ to get rid of the ${\log}_{2}$ function finding that $x = 64$

#### Explanation:

The first thing that we want to do is get rid of the logarithm on the left hand side of the equation. To do this, we want to perform an "anti-log" which is just a power. The inverse of a log function is the power of the same base, i.e.:

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

so in our case, we need to raise both sides of the equation to the exponent of $2$:

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

which simplifies to

$\frac{x}{2} = {2}^{5}$

now we can multiply both sides by $2$:

$x = 2 \cdot {2}^{5} = {2}^{1} \cdot {2}^{5} = {2}^{1 + 5} = {2}^{6} = 64$

Jul 15, 2016

$x = 64$

#### Explanation:

BY definition: if ${\log}_{a} b = c \text{ then } {a}^{c} = b$

In this case we have:

${\log}_{2} \left(\frac{x}{2}\right) = 5 \text{ } \Rightarrow {2}^{5} = \frac{x}{2}$

$\textcolor{m a \ge n t a}{2} \times {2}^{5} = \frac{x}{2} \times \textcolor{m a \ge n t a}{2}$

${2}^{6} = x$

$x = 64$