# How do you simplify log_2 (7^2 * 4^7)?

Mar 2, 2018

$2 {\log}_{2} 7 + 14$

#### Explanation:

You have to use the properties of logs.
Addition property: ${\log}_{b} \left(x y\right) = {\log}_{b} x + {\log}_{b} y$
Power property: ${\log}_{b} \left({x}^{y}\right) = y {\log}_{b} x$

By the addition property, ${\log}_{2} \left({7}^{2} \times {4}^{7}\right) = {\log}_{2} \left({7}^{2}\right) + {\log}_{2} \left({4}^{7}\right)$

By the power property, ${\log}_{2} \cdot \left({7}^{2}\right) = 2 {\log}_{2} 7$ and ${\log}_{2} \left({4}^{7}\right) = 7 {\log}_{2} 4$

So ${\log}_{2} \left({7}^{2} \times {4}^{7}\right) = 2 {\log}_{2} 7 + 7 {\log}_{2} 4$

${\log}_{2} 4 = 2$ so this evaluates as:

${\log}_{2} \left({7}^{2} \times {4}^{7}\right) = 2 {\log}_{2} 7 + 7 \left(2\right) = 2 {\log}_{2} 7 + 14$