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)(1/(log_x 2))=11 ⇒((11)/6)×(1/(log_x 2))=11 ⇒(1/(log_x 2))=11×(6/(11)) ⇒log_2 x=6 ⇒x=2^6 ∴x=64 is this rule correct????


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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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