The value of $x$ satisfying ${log}_{16} x + {log}_{x} 16 = {log}_{512} x + {log}_{x} 512$ is/are

\frac{1}{64}

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\frac{1}{16}

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64

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16

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Solution

The correct option is C $64$
Given
${log}_{16} x + {log}_{x} 16 = {log}_{512} x + {log}_{x} 512 \Rightarrow \frac{1}{4} {log}_{2} x + 4 {log}_{x} 2 = \frac{1}{9} {log}_{2} x + 9 {log}_{x} 2$
Assuming ${log}_{2} x = t$ , we get
$\Rightarrow \frac{t}{4} + \frac{4}{t} = \frac{t}{9} + \frac{9}{t} \Rightarrow \frac{t}{4} - \frac{t}{9} = \frac{9}{t} - \frac{4}{t} \Rightarrow \frac{5 t}{36} = \frac{5}{t} \Rightarrow t^{2} = 36 \Rightarrow t = \pm 6 \Rightarrow {log}_{2} x = \pm 6 \Rightarrow x = \frac{1}{64}, 64$

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