Question

If x² + y2 = 27xy, then show that log (x-y/5) =1/2 [ log x + log y]​

Answer 1

$\underline{\large{\sf To \:Prove :}}$

$\sf log (\frac{x-y}{5})=\frac{1}{2}[ log x + log y]$

$\underline{\large{\sf Proof :}}$

$\sf x^2+y^2=27xy$ --------------(Given)

$\sf x^2+y^2 = 25xy + 2xy$

$\sf x^2-2xy+y^2=25xy$

{from identity,$\sf(a-b)^2=a^2-2ab+b^2$}

∴ $\sf (x - y)^2 = 25xy$

∴ $\sf \frac{(x - y)^2}{25} = xy$

∴ $\sf (\frac{x-y}{5})^2 = xy$

taking square root on both the sides

$\sf (\frac{x-y}{5})=(xy)^{\frac{1}{2}}$

Now, taking log on both the sides

$\sf log (\frac{x-y}{5})=log(xy)^{\frac{1}{2}}$

we know, $\sf log_{e}( {m}^{n} ) = n log_{e}(m)$

∴ $\sf log (\frac{x-y}{5})=\frac{1}{2} log(xy)$

∴ $\sf log (\frac{x-y}{5}) =\frac{1}{2}(logx + logy)$........

...{since,$\sf log_{e}(mn) = log_{e}(m) + log_{e}(n)$}

therefore,

$\boxed{\sf log (\frac{x-y}{5})=\frac{1}{2}[ log x + log y]}$

Hence proved.