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Arithmetic Progression

If a^2+b...

Question

If $a^{2} + b^{2} = 7 a b$ , prove that $2 log (a + b) = log 9 + log a + log b$

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Solution

Given $a^{2} + b^{2} = 7 a b$
we have 2 log (a+b) = log 9 + log a + log b
now
using LHS
$2 l o g (a + b) = l o g (a + b)^{2}$
$\Rightarrow l o g [a^{2} + b^{2} + 2 a b]$
using given $\Rightarrow a^{2} + b^{2} = 7 a b$
$\Rightarrow l o g (7 a b + 2 a b) = l o g (9 a b)$
$\Rightarrow l o g 9 + l o g a + l o g b = R H S$

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Similar questions

a^{2} + b^{2} = 7 a b

, then prove that

log (\frac{a + b}{3}) = \frac{1}{2} (log a + log b)

a^{2} + b^{2} = 7 a b

log (\frac{1}{3} (a + b)) = \frac{1}{2} [log a + log b]

a^{2} - 12 a b + 4 b^{2} = 0

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

Q. Prove the following identities :
If

a^{2} + b^{2} = 7 a b

log \frac{1}{3} (a + b) = \frac{1}{2} [log a + log b]

a^{2} - 12 a b + {4 b}^{2} = 0,

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

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