Question

If log(x+y/2) = 1/2(log x+log y) show that x=y​
​

Answer 1

Given:-

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

To show:-

• x = y

Answer:-

Given that,

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

${\red{\bigstar}} \boxed{\sf{log\bigg(\dfrac{a}{b} \bigg) = log(a) - log(b)}}$

$\sf \longrightarrow \: log(x + y) - log(2) = \dfrac{1}{2} \bigg( log(x) + log(y) \bigg)$

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

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

${\green{\bigstar}} \boxed{\sf{blog(a) = log(a^b)}}$

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

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

${\red{\bigstar}} \boxed{\sf{log\bigg(\dfrac{a}{b}\bigg) = log(a) - log(b)}}$

$\sf \longrightarrow \: log \bigg( \dfrac{x + y}{ \sqrt{xy} } \bigg) = log(2)$

Comparing both sides,

$\sf \longrightarrow \dfrac{x + y}{ \sqrt{xy} } = 2$

$\sf \longrightarrow x + y = 2 \sqrt{xy}$

Squaring both sides,

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

$\sf \longrightarrow {x}^{2} + {y}^{2} + 2xy = 4xy$

$\sf \longrightarrow {x}^{2} + {y}^{2} - 2xy = 0$

$\sf \longrightarrow {(x - y)}^{2} = 0$

$\sf \longrightarrow (x - y)(x - y) = 0$

So,

$\longrightarrow \underline{ \underline{ \sf{ \green{ x = y }}}}$