Let x, y be real numbers such that x>2 y>0 and 2 log (x-2 y)=log x+log y Then, possible

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Basic of Logarithms

hard

Let $x, y$ be real numbers such that $x>2 y>0$ and $2 \log (x-2 y)=\log x+\log y$ Then, the possible value(s) of $\frac{x}{y}$

is $1$ only

are $1$ and $4$

is $4$ only

is $8$ only

(KVPY-2020)

(c)

Given, $\log (x-2 y)^2=\log x y$

$x^2+4 y^2-4 x y =x y$

$x^2+4 y^2-5 x y =0$

$\left(\frac{x}{y}\right)^2-5\left(\frac{x}{y}\right)+4 =0$

$\left(\frac{x}{y}-1\right)\left(\frac{x}{y}-4\right) =0 \Rightarrow \frac{x}{y}=1 \text { or } 4$

But if $\frac{x}{y}=1$ then, $x-2 y< 0$

Hence, only $\frac{x}{y}=4$

Standard 11

Mathematics

hard

normal

(KVPY-2017)

normal

easy

hard