Question

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Answer 1

Answer:-

This is the required solution.

Given,

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

$\sf \implies log( \frac{a - b}{5} ) = \frac{1}{2} log(ab)$

$\sf \implies log( \frac{a - b}{5} ) = log( \sqrt{ ab})$

Removing log from both side, we get,

$\sf \implies \frac{a - b}{5} = \sqrt{ ab}$

$\sf \implies a - b = 5\sqrt{ ab}$

Now, squaring both side, we get,

$\sf \implies (a - b)^{2} = (5\sqrt{ ab})^{2}$

$\sf \implies {a}^{2} + {b}^{2} - 2ab = 25 ab$

$\sf \implies {a}^{2} + {b}^{2} = 27 ab \: ...(i)$

Now,

$\sf \frac{a}{b} + \frac{b}{a}$

$\sf = \frac{ {a}^{2} + {b}^{2} }{ab}$

From (i), we can say that a²+b²=27 ab

So,

$\sf \frac{ {a}^{2} + {b}^{2} }{ab}$

$\sf = \frac{27 \cancel{ab}}{ \cancel{ab}}$

$\sf = 27$

Hence,

$\boxed{ \sf \frac{a}{b} + \frac{b}{a} = 27 }$

Which is the required answer.