If ${log}_{8} a + {log}_{4} b^{2} = 5, {log}_{8} b + {log}_{4} a^{2} = 7$ , then find the value of $a b$ .

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Solution

We know that
$l o g_{a^{n}} b = \frac{1}{n} l o g_{a} b$
$l o g_{a} b^{n} = n l o g_{a} b$
$l o g_{2} a + l o g_{2} b = l o g_{2} (a \times b)$

given $l o g_{8} a + l o g_{4} b^{2} = 5$
$l o g_{2^{3}} a + l o g_{2^{2}} b^{2} = 5$
$\Rightarrow \frac{1}{3} l o g_{2} a + 2 \times \frac{1}{2} l o g_{2} b = 5$

given $l o g_{8} b + l o g_{4} a^{2} = 5$
$l o g_{2^{3}} b + l o g_{2^{2}} a^{2} = 5$
$\Rightarrow \frac{1}{3} l o g_{2} b + 2 \times \frac{1}{2} l o g_{2} a = 7$

adding both, we get
$\frac{1}{3} l o g_{2} a + 2 \times \frac{1}{2} l o g_{2} b + \frac{1}{3} l o g_{2} b + 2 \times \frac{1}{2} l o g_{2} a = 7 + 5$
$\Rightarrow (\frac{1}{3} + 1) (l o g_{2} a + l o g_{2} b) = 12$
$\Rightarrow \frac{4}{3} (l o g_{2} (a \times b) = 12$
$l o g_{2} (a \times b) = 9$
$a \times b = 2^{9} = 512$

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