Question

if x2+y2,27xy then show that log (x-y)/5 =1/2.(logx+logy)​

Answer 1

❥︎Correct Question :-

If x² + y² = 27xy, prove that

$\bf \:log(\dfrac{x - y}{5} )= \dfrac{1}{2} ( log(x) + log(y) )$

❥︎Properties of log :-

$\bf \:logx + logy = logxy$

$\bf \:logx - logy = log\dfrac{x}{y}$

$\bf \:log {x}^{y} = ylogx$

❥︎Solution:-

$\bf \:x² + y² = 27xy$

Subtracting 2xy from both sides, we get

$\bf \:x² + y² - 2xy = 27xy - 2xy$

$\bf\implies \: {(x - y)}^{2} = 25xy$

Taking log on both sides, we get

$\bf\implies \:log {(x - y)}^{2} = log(25xy)$

$\bf\implies \:2log(x - y) = log25 + logx + logy$

$\bf\implies \:2log(x - y) = 2log5 + logx + logy$

$\bf\implies \:2log(x - y) - 2log5 = logx + logy$

$\bf\implies \:2(log(x - y) - log5) = logx + logy$

$\bf\implies \:log(\dfrac{x - y}{5} ) =\dfrac{1}{2} ( logx + logy)$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

_________________________________________