# How do you condense Logx+log2?

Mar 20, 2016

$\log \left(x\right) + \log \left(2\right) = \log \left(2 x\right)$

#### Explanation:

In general, ${\log}_{a} \left(x\right) + {\log}_{a} \left(y\right) = {\log}_{a} \left(x y\right)$ (see below for explanation).

Applying this, we get $\log \left(x\right) + \log \left(2\right) = \log \left(2 x\right)$

To see why the above property holds, note that ${\log}_{a} \left(x\right)$ is the value such that ${a}^{{\log}_{a} \left(x\right)} = x$. Then, as we have

${a}^{{\log}_{a} \left(x\right) + {\log}_{a} \left(y\right)} = {a}^{{\log}_{a} \left(x\right)} {a}^{{\log}_{a} \left(y\right)} = x y$

we see that ${\log}_{a} \left(x\right) + {\log}_{a} \left(y\right)$ is the value which, when taking $a$ to that power, results in $x y$. Thus we have

${\log}_{a} \left(x\right) + {\log}_{a} \left(y\right) = {\log}_{a} \left(x y\right)$