# How do you solve for x in log_2x-log_2 3=5?

$x = 96$

#### Explanation:

Let's first combine the left side, using the rule that ${\log}_{x} \left(a\right) - {\log}_{x} \left(b\right) = {\log}_{x} \left(\frac{a}{b}\right)$:

${\log}_{2} \left(x\right) - {\log}_{2} \left(3\right) = 5$

${\log}_{2} \left(\frac{x}{3}\right) = 5$

We can now set these as exponents to a base of 2 (to eliminate the log):

${2}^{{\log}_{2} \left(\frac{x}{3}\right)} = {2}^{5}$

On the left side, this has the effect of taking the inverse function of the log and so cancels out, leaving:

$\frac{x}{3} = 32$

$x = 96$

And now let's check it:

${\log}_{2} \left(96\right) - {\log}_{2} \left(3\right) = 5 \implies 6.585 - 1.585 = 5$

(the logs are rounded to 3 decimal places)