Question

If log --> = log Vox + los dos
show that (x+ y)2 = 20 xy​

Answer 1

Question : If $log( \frac{x - y}{4} ) = log( \sqrt{x} ) + log( \sqrt{y} )$

Show that, ${(x + y) }^{2} = 20xy$

Solution :

$log( \frac{x - y}{4} ) = log( \sqrt{x} ) + log( \sqrt{y} ) \\ \\ log( \frac{x - y}{4} ) = log( \sqrt{x} \times \sqrt{y} ) \\ \\ log( \frac{x - y}{4} ) = log( \sqrt{xy} ) \\ \\ ( \frac{x - y}{4}) = \sqrt{xy}$

Squaring on both sides,

$(x - y) ^{2} =( 4 \sqrt{xy} ) ^{2} \\ \\ {x}^{2} + {y}^{2} - 2xy = 16xy \\ \\ {x}^{2} + {y}^{2} = 18xy \\ \\ {x}^{2} + {y}^{2} + 2xy = 18xy + 2xy \\ \\ {(x + y)}^{2} = 20xy$

Hence proved that If $log( \frac{x - y}{4} ) = log( \sqrt{x} ) + log( \sqrt{y} )$

then, ${(x + y) }^{2} = 20xy$

Answer 2

Question :-

If {(x - y)/4} = log√x + log√y, show that (x + y)² = 20xy.

Solution :-

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

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

$\bf \fbox{ \because log \ a + log \ b = log \ ab}$

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

Cancelling log on both sides

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

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

Squaring on both sides

$\sf \implies (x - y)^2 = (4\sqrt{xy})^2$

$\sf \implies (x - y)^2 =16xy$

Expanding LHS

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

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

$\sf \implies x^2 + {y}^{2} =18xy$

Adding 2xy on both sides

$\sf \implies x^2 + {y}^{2} + 2xy=18xy + 2xy$

$\sf \implies (x + y)^2 =20xy$

$\bf \fbox{\because x^2 + {y}^{2} + 2xy= {(x + y)}^{2} }$

Hence shown