Question

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Answer 1

Required Answer:-

Given:

• $\sf \log \bigg( \dfrac{x + y}{4} \bigg) = \dfrac{1}{2}( \log(x) + \log(y))$

To prove:

• $\sf \dfrac{x}{y} + \dfrac{y}{x} = 14$

Solution:

We have,

$\sf \implies \log \bigg( \dfrac{x + y}{4} \bigg) = \dfrac{1}{2}( \log(x) + \log(y))$

We know that log(x) + log(y) = log(xy).

$\sf \implies \log \bigg( \dfrac{x + y}{4} \bigg) = \dfrac{1}{2}\log(xy)$

We also know that n log(x) = log(xⁿ)

$\sf \implies \log \bigg( \dfrac{x + y}{4} \bigg) = \log( \sqrt{xy} )$

Removing log from both sides, we get,

$\sf \implies \dfrac{x + y}{4} = \sqrt{xy}$

$\sf \implies x + y = 4\sqrt{xy}$

Squaring both sides, we get,

$\sf \implies {x}^{2} + {y}^{2} + 2xy = 16xy$

$\sf \implies {x}^{2} + {y}^{2} = 16xy - 2xy$

$\sf \implies {x}^{2} + {y}^{2} = 14xy$

Dividing both sides by xy, we get,

$\sf \implies \dfrac{{x}^{2}}{xy}+ \dfrac{{y}^{2}}{xy}= 14$

$\sf \implies \dfrac{x}{y}+ \dfrac{y}{x}= 14$

Hence Proved.