If log 24 x 2- x -1-log 2 x 2-1-1 then x lies in the intervalA- -2 - 3B- 1- -C- -2 - 3-0D- None of these

The correct options are A ( ∞, 2/3) B (1,∞) log_2(4x^2 x 1) log_2(x^2+1) gt;0 ⇒4x^2 x 1/x^2+1 gt;1 ⇒ 3x^2 x 2 gt;0 ⇒ (3x+2)(x 1) gt;0 ⇒ x lt; 2/3 ...

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Logarithms

If log 24 x 2...

Question

If $\frac{l o g_{2} (4 x^{2} - x - 1)}{l o g_{2} (x^{2} + 1)} > 1$ then x lies in the interval

$(- \infty, - 2 / 3)$

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$(1, \infty)$

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$(- 2 / 3, 0)$

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None of these

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If $\frac{l o g_{2} (4 x^{2} - x - 1)}{l o g_{2} (x^{2} + 1)} > 1$ then x lies in the interval

l o g_{\frac{1}{\sqrt{2}}} (x - 1) > 2

l o g_{\frac{1}{\sqrt{2}}} (x - 1) > 2

{log}_{\frac{1}{\sqrt{2}}} (x - 1) > 2,

x

{log}_{0.04} (x - 1) \geq {log}_{0.2} (x - 1),

x

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