The value of x satisfying log 16 x-log x 16-log 512 x-log x 512 is-are1A- 7-64B- 1-16C- 64D- 16

The correct options are A 164 C 64Given nbsp; log_16x+log_x16=log_512x+log_x512 ⇒ 1 4log_2 x+4log _x 2 = 19log _2x+9log_x nbsp; 2 Assuming log_2 x nbsp; = ...

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Byju's Answer

Standard XI

Mathematics

Fundamental Laws of Logarithms

The value of ...

Question

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 options are
A $\frac{1}{64}$
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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Similar questions

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The value of x satisfying log16x+logx16=log512x+logx512 is/are

The correct options are A 164 C 64Given log16x+logx16=log512x+logx512⇒14log2x+4logx2=19log2x+9logx2 Assuming log2x=t, we get ⇒t4+4t=t9+9t⇒t4−t9=9t−4t⇒5t36=5t⇒t2=36⇒t=±6⇒log2x=±6⇒x=164,64

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