# How do you solve x log 4 +log 8 = -x log 2 + log 10?

WARNING: This is not a logarithmic equation. In fact, the arguments of the logarithms are constants (numbers), not variables. So this is a simple linear equation with real (logarithmic) coefficients.

We can treat the coefficients as with integer or rational coefficients. We try to rearrange the equation to have all the terms that depend on $x$ on one side and "the numbers" on the other side:
$x \log 4 + x \log 2 = \log 10 - \log 8$
$x \left(\log 4 + \log 2\right) = \log 10 - \log 8$
$x = \frac{\log 10 - \log 8}{\log 4 + \log 2}$

Now we can use the following two properties of logarithms to simplify the expression:
$\log a + \log b = \log \left(a \cdot b\right)$
$\log c - \log d = \log \left(\frac{c}{d}\right)$
In our case $a = 4$, $b = 2$, $c = 10$ and $d = 8$, so $\log 4 + \log 2 = \log 6$ and $\log 10 - \log 8 = \log \left(\frac{10}{8}\right) = \log \left(\frac{5}{4}\right)$

We end with
$x = \log \frac{6}{\log} \left(\frac{5}{4}\right)$

Note
We can apply the change-of-base formula too:
$\log \frac{p}{\log} q = {\log}_{q} p$
The result will be written in an even more compact way, but the base of the logarithm would be quite unpractical. In fact, in our case $p = 6$ and $q = \frac{5}{4}$. We get

$x = {\log}_{\frac{5}{4}} 6$