Solve ${log}_{x} 2. {log}_{2 x} 2. {log}_{2} 4 x > 1$

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Solution

$l o g_{x} 2. l o g_{2 x} 2 . l o g_{2} 4 x > 1$

$\frac{1}{l o g_{2} x} . \frac{1}{l o g_{2} 2 x} . (l o g_{2} 4 + l o g_{2} x) > 1$

$\frac{1. (l o g_{2} 2^{2} + l o g_{2} x)}{l o g_{2} x . (l o g_{2} 2 + l o g_{2} x)} > 1$

If $x \neq 0, l o g_{2} x \neq - 1$ and if $x > 0, x \neq \frac{1}{2}$

$2 + l o g_{2} x > l o g_{2} x (l o g_{2} 2 + l o g_{2} x)$

$2 + l o g_{2} x > l o g_{2} x (1 + l o g_{2} x)$

$2 + l o g_{2} x > l o g_{2} x + l o g_{2}^{2} x$

$l o g_{2}^{2} x < 2$

$l o g_{2} x < \sqrt{2}$

$l o g_{2} x < l o g_{2} 2^{\sqrt{2}}$

$x < 2^{\sqrt{2}}$

But, $x = \frac{1}{2}$ makes the base 1, in the factor $l o g_{2 x} 2$ ; which is not allowed.
So, excluding the value $x = \frac{1}{2}$

$\Rightarrow x \in (0, \frac{1}{2}) \cup (\frac{1}{2}, 2^{\sqrt{2}})$

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{log}_{2} (4^{x + 1} + 4) \cdot {log}_{2} (4^{x} + 1) = {log}_{1 / \sqrt{2}} \sqrt{\frac{1}{8}}

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

log 2 (4^{x + 1} + 4) log 2 (4^{x} + 1) = log \frac{1}{\sqrt{2}} \sqrt{\frac{1}{8}}

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

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