# How do you solve  log_2(24) - log_2(3) = log_2x?

Dec 3, 2015

$x = 8$

#### Explanation:

Use the logarithmic law

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

Thus,

${\log}_{2} \left(24\right) - {\log}_{2} \left(3\right) = {\log}_{2} \left(x\right)$

$\iff {\log}_{2} \left(\frac{24}{3}\right) = {\log}_{2} \left(x\right)$

$\iff {\log}_{2} \left(8\right) = {\log}_{2} \left(x\right)$

At this point, you already see that $x = 8$.

To make it even more clear, let's get rid of the logarithmic expressions.

The inverse function of ${\log}_{2} \left(x\right)$ is ${2}^{x}$ which means that ${\log}_{2} \left({2}^{x}\right) = x$ and also ${2}^{{\log}_{2} \left(x\right)} = x$ hold.

This means:

${\log}_{2} \left(8\right) = {\log}_{2} \left(x\right)$

$\iff {2}^{{\log}_{2} \left(8\right)} = {2}^{{\log}_{2} \left(x\right)}$

$\iff 8 = x$