Question

Xpower2+y power2=27xy
find the value of log (x-y/5)=1/a(logx+logy)

Answer 1

$Given \: x^{2} + y^{2} = 27xy$

/* Subtract bothsides by 2xy , we get */

$\implies x^{2} + y^{2} - 2xy = 27xy - 2xy$

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

$\implies (x-y)^{2} = 5^{2}xy$

/* Dividing both sides by 5², we get */

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

$\implies \Big(\frac{x-y}{5}\Big)^{2} = xy$

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

$\implies 2log \Big(\frac{x-y}{5}\Big)= log x + log y$

$\boxed {\pink { log a^{m} = m log a }}$

$\implies log \Big(\frac{x-y}{5}\Big)= \frac{1}{2}( log x + log y)$

$Hence \:proved$

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