Question

please answer the questions..................​

Answer 1

Answer:

$\mathfrak{\huge{\orange{\underline{\underline{QuEsTiOn :}}}}}$

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

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

$\Large{\boxed{\underline{\overline{\mathfrak{\star \: AnSwer :- \: \star}}}}}$

$\Large{\underline{\underline{\bf{SoLuTion:-}}}}$

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

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