# How do you solve log_5(2x+15)=log_5(3x)?

$x = 15$
We start with ${\log}_{5} \left(2 x + 15\right) = {\log}_{5} \left(3 x\right)$. In this case we got lucky and got an equation that has the same base, $5$. We can use this to our adantage, as long as we know how to convert a logarithm to an exponent. We do that using this rule: ${\log}_{\textcolor{red}{b}} \textcolor{b l u e}{x} = \textcolor{g r e e n}{y}$ becomes ${\textcolor{red}{b}}^{\textcolor{g r e e n}{y}} = \textcolor{b l u e}{x}$. The base stays the same and $x$ and $y$ switch.
Let's set up our conversion: ${\log}_{\textcolor{red}{5}} \textcolor{b l u e}{2 x + 15} = \textcolor{g r e e n}{{\log}_{5} \left(3 x\right)}$ becomes ${\textcolor{red}{5}}^{\textcolor{g r e e n}{{\log}_{5} \left(3 x\right)}} = \textcolor{b l u e}{2 x + 15}$. Now that we've got this set up, let's clean it up. Notice the ${\textcolor{red}{5}}^{\textcolor{g r e e n}{{\log}_{5}}}$ part. These two are inverses of eachother, so they cance out. That leaves us with ${\cancel{\textcolor{red}{5}}}^{\cancel{\textcolor{g r e e n}{{\log}_{5}}}} {\textcolor{w h i t e}{1}}^{\textcolor{g r e e n}{3 x}} = \textcolor{b l u e}{2 x + 15}$. The $3 x$ dops down and gives us $3 x = 2 x + 15$. That can be simplifed by subtracting $2 x$ on both sides, which leaves us with $x = 15$.