Question

If log(x+y/3)= log x + logy, then
find the value of x/y+ y/x​

Answer 1

Question:-

• If $\sf \log(\frac{x+y}{3}) = \log(x)+\log(y)$, then find the value of $\sf \frac{x}{y}+\frac{y}{x}$

Answer:-

• $\boxed{ \sf \frac{x}{y} + \frac{y}{x} = 9xy - 2}$

Solution:-

Given,

$\sf \log( \frac{x + y}{3} ) = \log(x) + \log(y)$

$\sf \implies \log( \frac{x + y}{3} ) = \log(xy)$

Now, removing log from both the sides, we get,

$\sf \implies \frac{x + y}{3} = xy$

$\sf \implies x + y= 3xy$

Now, squaring both sides, we get,

$\sf \implies {(x + y)}^{2} = (3xy)^{2}$

$\sf \implies {x}^{2} + {y}^{2} + 2xy = 9 {x}^{2} {y}^{2}$

$\sf \implies {x}^{2} + {y}^{2}= 9 {x}^{2} {y}^{2} - 2xy$

$\sf \implies {x}^{2} + {y}^{2}=xy(9xy - 2)$

Now,

$\sf \frac{x}{y} + \frac{y}{x}$

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

$\sf = \frac{ \cancel{xy}(9xy - 2)}{ \cancel{xy} }$

$\sf = 9xy - 2$

Formula Used:-

• $\sf\log(a)+\log(b)+\log(c)+...=\log(abc...)$