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If log a-b/2...

Question

If $l o g \frac{a - b}{2} = \frac{1}{2} (l o g a + l o g b)$ , show that:
$a^{2} + b^{2} = 6 a b$

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Solution

log $open square brackets fraction numerator a minus b over denominator 2 end fraction close square brackets$ =(logab)
log $open square brackets fraction numerator a minus b over denominator 2 end fraction close square brackets$ =log
$open square brackets fraction numerator a minus b over denominator 2 end fraction close square brackets$ =
$open square brackets fraction numerator a minus b over denominator 2 end fraction close square brackets squared$ =ab
=ab
=4ab
-2ab=4ab
=6ab

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Similar questions

log (\frac{a - b}{2}) = \frac{1}{2} (log a + log b)

, then the value of

a^{2} + b^{2} - 6 a b

a^{2} + b^{2} = 23 a b

log \frac{a + b}{5} = \frac{1}{2} (log a + log b)

a^{2} + b^{2} = 3 a b,

log (\frac{a + b}{\sqrt{5}}) = \frac{1}{2} (log a + log b)

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