Question

x2 +y2=27xy then show that log [x-y÷5]=[log x- log y]​

Answer 1

Step-by-step explanation:

ToProve:

\sf log (\frac{x-y}{5})=\frac{1}{2}[ log x + log y]log(5x−y)=21[logx+logy]

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

\sf x^2+y^2=27xyx2+y2=27xy --------------(Given)

\sf x^2+y^2 = 25xy + 2xyx2+y2=25xy+2xy

\sf x^2-2xy+y^2=25xyx2−2xy+y2=25xy

{from identity,\sf(a-b)^2=a^2-2ab+b^2(a−b)2=a2−2ab+b2 }

∴ \sf (x - y)^2 = 25xy(x−y)2=25xy

∴ \sf \frac{(x - y)^2}{25} = xy25(x−y)2=xy

∴ \sf (\frac{x-y}{5})^2 = xy(5x−y)2=xy

taking square root on both the sides

\sf (\frac{x-y}{5})=(xy)^{\frac{1}{2}}(5x−y)=(xy)21

Now, taking log on both the sides

\sf log (\frac{x-y}{5})=log(xy)^{\frac{1}{2}}log(5x−y)=log(xy)21

we know, \sf log_{e}( {m}^{n} ) = n log_{e}(m)loge(mn)=nloge(m)

∴ \sf log (\frac{x-y}{5})=\frac{1}{2} log(xy)log(5x−y)=21log(xy)

∴ \sf log (\frac{x-y}{5}) =\frac{1}{2}(logx + logy)log(5x−y)=21(logx+logy) ........

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

therefore,

\boxed{\sf log (\frac{x-y}{5})=\frac{1}{2}[ log x + log y]}log(5x−y)=21[logx+logy]

Hence proved.