The solution set of the equation ${log}_{x} 2. {log}_{2 x} 2 = {log}_{4 x} 2$ is

2

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

2^{- \sqrt{2}}

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

2^{\sqrt{2}}

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

\frac{1}{2}

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

Open in App

Solution

The correct options are
B $2^{- \sqrt{2}}$
C $2^{\sqrt{2}}$
Given, $l o g_{x}^{2} . l o g_{2 x}^{2} . l o g_{2 x}^{2} = l o g_{4 x}^{2}$

$\Rightarrow \frac{1}{l o g_{2}^{x}} . \frac{1}{l o g_{2}^{2 x}} = \frac{1}{l o g_{2} 4 x}$ $[∵ l o g_{a}^{x} = \frac{1}{l o g_{x}^{a}}]$

$l o g_{2}^{x} (l o g_{2}^{2} + l o g_{2}^{x}) = l o g_{2}^{4} + l o g_{2}^{x}$ $[∵ l o g a b = l o g a + l o g b]$

$l o g_{2}^{x} + (l o g_{2} x)^{2} = 2 + l o g_{2}^{x}$

$l o g_{2}^{x} = \pm \sqrt{2}$

$x = 2^{\pm \sqrt{2}}$

$∴$ Option B,C are both correct

Suggest Corrections

Similar questions

{log}_{x} 2. {log}_{2 x} 2 = {log}_{4 x} 2

{log}_{x} 2 \cdot {log}_{2 x} 2 = {log}_{4 x} 2

A

{log}_{x} 2 \cdot {log}_{2 x} 2 = {log}_{4 x} 2

B

x^{{log}_{x} (3 - x)^{2}} = 25,

n (A \cup B)

2 - {log}_{2} (x^{2} + 3 x) \geq 0

{log}_{2 x} 2 \cdot {log}_{x} 2 = {log}_{4 x} 2,

x

Join BYJU'S Learning Program