How do you solve $log (y) = \frac{1}{4} log 16 + \frac{1}{2} log 49$ $log(y) = 1/4 log16 + 1/2 log49$ ?

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How do you solve $log (y) = \frac{1}{4} log 16 + \frac{1}{2} log 49$ $log(y) = 1/4 log16 + 1/2 log49$ ?

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Answer 1 · 2015-08-02T07:41:21

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

So $y = 14$ $y = 14$

Explanation:

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

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

So:

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

$= \frac{1}{4} \cdot 4 log 2 + \frac{1}{2} \cdot 2 log 7 = log 2 + log 7 = log (2 \cdot 7)$ $= 1/4*4 log 2 + 1/2 * 2 log 7 = log 2 + log 7 = log (2*7)$

$= log 14$ $= log 14$

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

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

So $y = 14$ $y = 14$

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