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

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Answer 1 · 2015-08-02T07:41:21

$log y = 1/4 log 16 + 1/2 log 49 = log 2 + log 7 = log 14$

So $y = 14$

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

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

So:

$log y = 1/4 log 16 + 1/2 log 49 = 1/4 log 2^4 + 1/2 log 7^2$

$= 1/4*4 log 2 + 1/2 * 2 log 7 = log 2 + log 7 = log (2*7)$

$= log 14$

That is: $log y = log 14$

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

So $y = 14$

How do you solve log(y) = 1/4 log16 + 1/2 log49log(y) = 1/4 log16 + 1/2 log49?