Question

if 2 log (X+y/4)=logx+logy ,then find the value of X/y+y/X?​

Answer 1

We have:

$2\log (\dfrac{x+y}{4})=\log x+\log y$

We have to find the value of $\dfrac{x}{y} +\dfrac{y}{x}$.

Solution:

∴ $2\log (\dfrac{x+y}{4})=\log x+\log y$

Using logarithm identity:

$b\log a=\log a^b$ and $\log x+\log y=\log xy$

⇒ $\log (\dfrac{x+y}{4})^2=\log xy$

⇒ $(\dfrac{x+y}{4})^2=xy$

⇒ $(x+y)^2=16xy$

⇒ $x^2+y^2+2xy=16xy$

⇒ $x^2+y^2=14xy$

Dividing both sides by xy, we get

$\dfrac{x^2+y^2}{xy} =\dfrac{14xy}{xy}$

⇒ $\dfrac{x}{y} +\dfrac{y}{x}=14$

Thus, the value of $\dfrac{x}{y} +\dfrac{y}{x}$ is equal to "14".

Answer 2

Answer:

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