Question Number 194613 by cortano12 last updated on 11/Jul/23
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Answered by MM42 last updated on 11/Jul/23
$$\frac{\mathrm{1}}{{log}_{{x}} \mathrm{4}{x}}+\frac{\mathrm{1}}{{log}_{{x}} \frac{{x}}{\mathrm{2}}}=\mathrm{2} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{log}_{{x}} \mathrm{2}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}−{log}_{{x}} \mathrm{2}}=\mathrm{2} \\ $$$${let}\:\:{log}_{{x}} \mathrm{2}={y} \\ $$$$\mathrm{4}{y}^{\mathrm{2}} −{y}=\mathrm{0}\Rightarrow{y}=\mathrm{0}={log}_{{x}} \mathrm{2}\:\:× \\ $$$${or}\:\:{y}=\frac{\mathrm{1}}{\mathrm{4}}={log}_{{x}} \mathrm{2}\Rightarrow{x}=\mathrm{16}\:\checkmark \\ $$$$ \\ $$
Answered by mahdipoor last updated on 11/Jul/23
$${log}_{{ab}} {c}=\frac{\mathrm{1}}{{log}_{{c}} {ab}}=\frac{\mathrm{1}}{{log}_{{c}} {a}+{log}_{{c}} {b}} \\ $$$${log}_{\mathrm{4}{x}} {x}+{log}_{{x}/\mathrm{2}} {x}=\frac{\mathrm{1}}{{log}_{{x}} \mathrm{4}+{log}_{{x}} {x}}+\frac{\mathrm{1}}{{log}_{{x}} {x}+{log}_{{x}} \mathrm{1}/\mathrm{2}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{log}_{{x}} \mathrm{2}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}−{log}_{{x}} \mathrm{2}}=\mathrm{2}\Rightarrow{log}_{{x}} \mathrm{2}={u}\neq\mathrm{1},−.\mathrm{5}\Rightarrow \\ $$$$\left(\mathrm{1}−{u}\right)+\left(\mathrm{2}{u}+\mathrm{1}\right)=\mathrm{2}\left(\mathrm{1}−{u}\right)\left(\mathrm{2}{u}+\mathrm{1}\right)\Rightarrow \\ $$$${log}_{{x}} \mathrm{2}=\frac{\mathrm{1}}{{log}_{\mathrm{2}} {x}}=\frac{\mathrm{1}}{\mathrm{4}},\mathrm{0}\Rightarrow{x}=\mathrm{2}^{\mathrm{4}} =\mathrm{16} \\ $$