Solution-log-8-x-log-4-x-log-2-x-11-1-log-x-8-1-log-x-4-1-log-x-2-11-1-log-x-2-3-1-log-x-2-2-1-log-x-2-11-1-3log-x-2-1-2log-x-2-1-log-x-2-11-1-3-1-2-1-

Question Number 70017 by Shamim last updated on 30/Sep/19

Solution -

\log_{8} x + \log_{4} x + \log_{2} x = 11

\Rightarrow \frac{1}{\log_{x} 8} + \frac{1}{\log_{x} 4} + \frac{1}{\log_{x} 2} = 11

\Rightarrow \frac{1}{\log_{x} 2^{3}} + \frac{1}{\log_{x} 2^{2}} + \frac{1}{\log_{x} 2} = 11

\Rightarrow \frac{1}{{3 \log}_{x} 2} + \frac{1}{{2 \log}_{x} 2} + \frac{1}{\log_{x} 2} = 11

\Rightarrow (\frac{1}{3} + \frac{1}{2} + 1) \frac{1}{\log_{x} 2} = 11

\Rightarrow \frac{11}{6} \times \frac{1}{\log_{x} 2} = 11

\Rightarrow \frac{1}{\log_{x} 2} = 11 \times \frac{6}{11}

\Rightarrow \log_{2} x = 6

\Rightarrow x = 2^{6}

∴ x = 64

is this rule correct ? ? ? ?

Commented by Tony Lin last updated on 30/Sep/19

c o r r e c t!

Commented by Shamim last updated on 30/Sep/19

tnks to u .

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