Question Number 161786 by mathlove last updated on 22/Dec/21
$${log}\underset{\mathrm{4}{x}} {{x}}+{log}\underset{\frac{{x}}{\mathrm{2}}} {{x}}=\mathrm{2} \\ $$$${solve}\:\:\:{for}\:\:\:{x}=? \\ $$
Answered by mr W last updated on 22/Dec/21
$$\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\left(\mathrm{4}{x}\right)}+\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\left(\frac{{x}}{\mathrm{2}}\right)}=\mathrm{2} \\ $$$$\frac{\mathrm{ln}\:{x}}{\mathrm{2ln}\:\mathrm{2}+\mathrm{ln}\:{x}}+\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:{x}−\mathrm{ln}\:\mathrm{2}}=\mathrm{2} \\ $$$${let}\:{t}=\mathrm{ln}\:{x},\:{a}=\mathrm{ln}\:\mathrm{2} \\ $$$$\frac{{t}}{\mathrm{2}{a}+{t}}+\frac{{t}}{{t}−{a}}=\mathrm{2} \\ $$$${t}=\mathrm{4}{a} \\ $$$$\mathrm{ln}\:{x}=\mathrm{4}\:\mathrm{ln}\:\mathrm{2} \\ $$$$\Rightarrow{x}=\mathrm{2}^{\mathrm{4}} =\mathrm{16} \\ $$
Answered by cortano last updated on 22/Dec/21
$$\:\left({Q}\right)\:{Solve}\:{for}\:{x}\::\:\mathrm{log}\:_{\mathrm{4}{x}} \left({x}\right)+\mathrm{log}\:_{\frac{{x}}{\mathrm{2}}} \left({x}\right)\:=\:\mathrm{2} \\ $$$$\:\left({S}\right)\:\Rightarrow\:\frac{\mathrm{log}\:_{\mathrm{2}} \left({x}\right)}{\mathrm{2}+\mathrm{log}\:_{\mathrm{2}} \left({x}\right)}\:+\:\frac{\mathrm{log}\:_{\mathrm{2}} \left({x}\right)}{\mathrm{log}\:_{\mathrm{2}} \left({x}\right)−\mathrm{1}}\:=\:\mathrm{2} \\ $$$$\:\left[\:{z}\:=\:\mathrm{log}\:_{\mathrm{2}} \left({x}\right);\:{x}>\mathrm{0}\:;\:{x}\neq\mathrm{1}\:\right] \\ $$$$\:\Rightarrow\:\frac{{z}}{{z}+\mathrm{2}}\:+\:\frac{{z}}{{z}−\mathrm{1}}\:=\:\mathrm{2} \\ $$$$\Rightarrow{z}\left({z}−\mathrm{1}+{z}+\mathrm{2}\right)=\:\mathrm{2}\left({z}+\mathrm{2}\right)\left({z}−\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{2}{z}^{\mathrm{2}} +{z}\:=\:\mathrm{2}{z}^{\mathrm{2}} \:+\mathrm{2}{z}−\mathrm{4} \\ $$$$\:\Rightarrow{z}\:=\:\mathrm{4}=\:\mathrm{log}\:_{\mathrm{2}} \left({x}\right) \\ $$$$\Rightarrow{x}\:=\:\mathrm{2}^{\mathrm{4}} =\:\mathrm{16} \\ $$