# How do I solve log_2 7 = log_2 (7/16)?

## I'm not sure if what I'm doing is correct, but I tried expanding it: ${\log}_{2} 7 - {\log}_{2} 7 - {\log}_{2} 16$ $= {\log}_{2} \left(\frac{7}{7}\right) - {\log}_{2} 16$ $= {\log}_{2} 1 - {\log}_{2} 16$ $= {\log}_{2} \left(\frac{1}{16}\right)$ $= {\log}_{2} \left(\frac{1}{2} ^ 4\right)$ and after that I just got stuck. Can someone help?

Oct 15, 2017

The equation is the same as 4.

#### Explanation:

I'm guessing that based on your description that it is meant to be ${\log}_{2} \left(7\right) - {\log}_{2} \left(\frac{7}{16}\right)$ and not ${\log}_{2} \left(7\right) = {\log}_{2} \left(\frac{7}{16}\right)$

There are two ways of doing this.

The way you did it is almost correct except you forgot to multiply by -1 after expanding $- {\log}_{2} \left(\frac{7}{16}\right)$.

Remember, you are taking the negative of ${\log}_{2} \left(\frac{7}{16}\right)$. The positive of ${\log}_{2} \left(\frac{7}{16}\right)$ is ${\log}_{2} \left(7\right) - {\log}_{2} \left(16\right)$, while the negative is $- \left({\log}_{2} \left(7\right) - {\log}_{2} \left(16\right)\right) = - {\log}_{2} \left(7\right) + {\log}_{2} \left(16\right)$

${\log}_{2} \left(7\right) - {\log}_{2} \left(\frac{7}{16}\right)$
${\log}_{2} \left(7\right) - \left({\log}_{2} \left(7\right) - {\log}_{2} \left(16\right)\right)$
${\log}_{2} \left(7\right) - {\log}_{2} \left(7\right) + {\log}_{2} \left(16\right)$
${\log}_{2} \left(\frac{7}{7}\right) + {\log}_{2} \left(16\right)$
${\log}_{2} \left(1\right) + {\log}_{2} \left(16\right)$
$0 + {\log}_{2} \left(16\right)$
$= {\log}_{2} \left(16\right) = 4$

You know that ${\log}_{a} b - {\log}_{a} c = {\log}_{a} \left(\frac{b}{c}\right)$

Well, ${\log}_{2} \left(7\right) - {\log}_{2} \left(\frac{7}{16}\right)$
$= {\log}_{2} \left(\frac{7}{\frac{7}{16}}\right)$
$= {\log}_{2} \left(\frac{\cancel{7} \cdot 16}{\cancel{7}}\right)$
$= {\log}_{2} \left(16\right) = 4$