Question

If x² + y^2 = 51xy, prove that
log x-y/7 = 1/2(log x + log y).​

Answer 1

Given,

$\sf{x^2+y^2=51xy}$

Subtracting $\sf{2xy}$ from both sides,

$\sf{x^2-2xy+y^2=49xy}\\\\\\\sf{(x-y)^2=49xy}$

Taking logarithm on both sides,

$\sf{\log (x-y)^2=\log(49xy)}$

We know that,

$\sf{\log a^b=b\log a}\\\\\\\sf{\log (abc)=\log a+\log b+\log c}$

Then,

$\sf{2\log (x-y)=\log 49+\log x+\log y}\\\\\\\sf{2\log (x-y)=\log \left(7^2\right)+\log x+\log y}\\\\\\\sf{2\log (x-y)=2\log7+\log x+\log y}\\\\\\\sf{2\log (x-y)-2\log 7=\log x+\log y}\\\\\\\sf{2\left [\log (x-y)-\log 7\right]=\log x+\log y}$

But,

$\sf{\log a-\log b=\log \left (\dfrac {a}{b}\right)}$

Then,

$\sf{2\log \left(\dfrac {x-y}{7}\right)=\log x+\log y}$

Finally,

$\sf{\log \left(\dfrac {x-y}{7}\right)=\dfrac {1}{2}\left (\log x+\log y\right)}$

Hence Proved!