# How do you solve log_2(4x)=log_2(x+15)?

Mar 16, 2018

Since both sides are taking the same logarithm, you can directly compare the terms inside the parentheses and find that $x = 5$

#### Explanation:

The easy way to look at this is by realizing that you can compare the in-parentheses terms directly. This is because they're both logarithms of the same base.

A base-2 log has equivalence like this:

${\log}_{2} \left(4 x\right) = n \Rightarrow {2}^{n} = 4 x$

So if:

${2}^{n} = 4 x$

and

${2}^{n} = x + 15$

then

$4 x = x + 15$

Now, we can solve for $x$:

$3 x = 15 \Rightarrow \textcolor{red}{x = 5}$

Mar 16, 2018

$x = 5$

#### Explanation:

${\log}_{2} \left(4 x\right) = {\log}_{2} \left(x + 15\right)$

Or, $1 = \frac{{\log}_{2} \left(x + 15\right)}{{\log}_{2} \left(4 x\right)} = {\log}_{4 x} \left(x + 15\right)$

So,${\left(4 x\right)}^{1} = x + 15$

Or, $3 x = 15$

So,$x = 5$