Question

16) If x2+y2=7xythen show that2log(x+y divide by 3)=logx +logy​

Answer 1

Given that,

${x}^{2} + {y}^{2} = 7xy$

Adding 2xy to both sides,

$\longrightarrow {x}^{2} + {y}^{2} + 2xy = 7xy + 2xy$

$\longrightarrow {(x + y)}^{2} = 9xy$

$\longrightarrow {(x + y)}^{2} = {(3 \sqrt{xy})}^{2}$

Taking square root on both sides,

$\longrightarrow \sqrt{{(x + y)}^{2}} = \sqrt{{(3 \sqrt{xy})}^{2}}$

Considering only positive values,

$\longrightarrow x + y = 3 \sqrt{xy}$

$\longrightarrow \sqrt{xy} = \dfrac{x + y}{3} \quad \quad \dots(1)$

Now, coming to the required proof,

$2 \log \bigg( \dfrac{x + y}{3} \bigg) = \log(x) + \log(y)$

Putting the value of (x + y)/3 from (1) in L.H.S.,

$\longrightarrow 2 \log ( \sqrt{xy} ) = \log(x) + \log(y)$

$\longrightarrow 2 \log \{({xy})^{1/2} \} = \log(x) + \log(y)$

$\red{ \bigstar} \boxed{ \log(a) {}^{b} = b \log(a) }$

$\longrightarrow 2 \times \dfrac{1}{2} \log (xy) = \log(x) + \log(y)$

$\longrightarrow \log (xy) = \log(x) + \log(y)$

$\green{ \bigstar} \boxed{ \log(ab) = \log(a) + \log(b)}$

$\longrightarrow \log (xy) = \log(xy)$

Hence proved!