Suppose x, y are positive real numbers such that 2 log (x - 2y) = log x + logy. Then value of x/y is

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3/2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2/3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
4

2log(x - 2y) = logx + logy

log(x - 2y)² = log(xy)

(x - 2y)²= xy

x² + 4y² - 4xy = xy

x²+ 4y² - 5xy = 0

x²- 4xy - xy + 4y² = 0

x(x - 4y) - y (x - 4y) = 0

(x - y)(x - 4y) = 0

x = y (Not possible), x = 4y

x/y = 4

Suggest Corrections

Similar questions

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & l o g_{x} y & l o g_{x} z l o g_{y} x & 1 & l o g_{y} z l o g_{z} x & l o g_{z} y & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

x

y

x \sqrt{x} + y \sqrt{y} = 183

x \sqrt{y} + y \sqrt{x} = 182

\frac{18}{5} (x + y)

x

1

{log}_{a} x, {log}_{b} x, {log}_{c} x

{log}_{10} (x^{3} + y^{3}) - {log}_{10} (x^{2} + y^{2} - x y) \leq 2

x y

Join BYJU'S Learning Program