# How do you express 3log_5 4 - 5log_5 2 as a single logarithm?

Apr 8, 2017

$3 {\log}_{5} 4 - 5 {\log}_{5} 2 = \frac{1}{\log} _ 2 5$

#### Explanation:

To solve this, you will need to use the "Change of Base" formula:
${\log}_{a} x = \frac{{\log}_{b} x}{{\log}_{b} a}$
where we change the base-$a$ logarithm to base-$b$.

Let's see how we can apply this to what we have.
I notice that we have a 4 and a 2 inside a base-5 logarithm.
So maybe I should change to a base-2 logarithm.
Let's see how each log transforms:
${\log}_{5} 4 = \frac{{\log}_{2} 4}{{\log}_{2} 5}$
but since $4 = {2}^{2}$,
${\log}_{2} 4 = {\log}_{2} {2}^{2} = 2$
so
${\log}_{5} 4 = \frac{2}{{\log}_{2} 5}$
and similarly,
${\log}_{5} 2 = \frac{1}{{\log}_{2} 5}$

So, our expression becomes:
$3 {\log}_{5} 4 - 5 {\log}_{5} 2 = \frac{3 \cdot 2}{{\log}_{2} 5} - \frac{5 \cdot 1}{{\log}_{2} 5}$
i.e.
$= \frac{6 - 5}{{\log}_{2} 5}$
$= \frac{1}{\log} _ 2 5$

There could be other answers using different bases.