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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  ⇒(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????
$$\mathrm{Solution}- \\ $$$$\mathrm{log}_{\mathrm{8}} \mathrm{x}+\mathrm{log}_{\mathrm{4}} \mathrm{x}+\mathrm{log}_{\mathrm{2}} \mathrm{x}=\mathrm{11} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{8}}+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{4}}+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}}=\mathrm{11} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}}=\mathrm{11} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3log}_{\mathrm{x}} \mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2log}_{\mathrm{x}} \mathrm{2}}+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}}=\mathrm{11} \\ $$$$\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}}=\mathrm{11} \\ $$$$\Rightarrow\frac{\mathrm{11}}{\mathrm{6}}×\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}}=\mathrm{11} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}}=\mathrm{11}×\frac{\mathrm{6}}{\mathrm{11}} \\ $$$$\Rightarrow\mathrm{log}_{\mathrm{2}} \mathrm{x}=\mathrm{6} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{2}^{\mathrm{6}} \\ $$$$\therefore\mathrm{x}=\mathrm{64} \\ $$$$ \\ $$$$\mathrm{is}\:\mathrm{this}\:\mathrm{rule}\:\mathrm{correct}???? \\ $$
Commented by Tony Lin last updated on 30/Sep/19
correct!
$${correct}! \\ $$
Commented by Shamim last updated on 30/Sep/19
tnks to u.
$$\mathrm{tnks}\:\mathrm{to}\:\mathrm{u}. \\ $$

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