# How do you solve log_2 (3x)-log_2 7=3?

Jul 8, 2016

Use a property of logs to simplify and solve an algebraic equation to get $x = \frac{56}{3}$.

#### Explanation:

Begin by simplifying ${\log}_{2} 3 x - {\log}_{2} 7$ using the following property of logs:
$\log a - \log b = \log \left(\frac{a}{b}\right)$
Note that this property works with logs of every base, including $2$.

Therefore, ${\log}_{2} 3 x - {\log}_{2} 7$ becomes ${\log}_{2} \left(\frac{3 x}{7}\right)$. The problem now reads:
${\log}_{2} \left(\frac{3 x}{7}\right) = 3$

We want to get rid of the logarithm, and we do that by raising both sides to the power of $2$:
${\log}_{2} \left(\frac{3 x}{7}\right) = 3$
$\to {2}^{{\log}_{2} \left(\frac{3 x}{7}\right)} = {2}^{3}$
$\to \frac{3 x}{7} = 8$

Now we just have to solve this equation for $x$:
$\frac{3 x}{7} = 8$
$\to 3 x = 56$
$\to x = \frac{56}{3}$

Since this fraction cannot be simplified further, it is our final answer.