# How do you solve log(y) = 1/4 log16 + 1/2 log49?

Aug 2, 2015

$\log y = \frac{1}{4} \log 16 + \frac{1}{2} \log 49 = \log 2 + \log 7 = \log 14$

So $y = 14$

#### Explanation:

If $a > 0$ then $\log {a}^{b} = b \log a$

If $a , b > 0$ then $\log a b = \log a + \log b$

So:

$\log y = \frac{1}{4} \log 16 + \frac{1}{2} \log 49 = \frac{1}{4} \log {2}^{4} + \frac{1}{2} \log {7}^{2}$

$= \frac{1}{4} \cdot 4 \log 2 + \frac{1}{2} \cdot 2 \log 7 = \log 2 + \log 7 = \log \left(2 \cdot 7\right)$

$= \log 14$

That is: $\log y = \log 14$

If $\log a = \log b$ then $a = b$

So $y = 14$