If log24x2-x-1-log2x2-1-1 then x lies in the interval

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 or x gt;1 ...

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Logarithms

If log24x2-x-...

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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

(- \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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