Question Number 158593 by mathlove last updated on 06/Nov/21
Answered by cortano last updated on 07/Nov/21
![{ ((log _y (x)=log _x (2))),((log _x (y).log _x (y)=log _x (16))) :} ⇒[ log _x (y)]^2 =4log _x (2) let log _x (y)=u⇒log _y (x)=(1/u) ⇒u^2 =(4/u) ⇒u^3 =4 ⇒u=(4)^(1/3) ⇒log _x (y)=(4)^(1/3) ⇒y=x^(4)^(1/3) ⇒log_x^(4)^(1/3) (x)=log_x (2) ⇒(1/( (4)^(1/3) )) log_x (x)=log_x (2) ⇒2^(4)^(1/3) =x](Q158594.png)
$$\begin{cases}{\mathrm{log}\:_{{y}} \left({x}\right)=\mathrm{log}\:_{{x}} \left(\mathrm{2}\right)}\\{\mathrm{log}\:_{{x}} \left({y}\right).\mathrm{log}\:_{{x}} \left({y}\right)=\mathrm{log}\:_{{x}} \left(\mathrm{16}\right)}\end{cases} \\ $$$$\Rightarrow\left[\:\mathrm{log}\:_{{x}} \left({y}\right)\right]^{\mathrm{2}} =\mathrm{4log}\:_{{x}} \left(\mathrm{2}\right) \\ $$$${let}\:\mathrm{log}\:_{{x}} \left({y}\right)={u}\Rightarrow\mathrm{log}\:_{{y}} \left({x}\right)=\frac{\mathrm{1}}{{u}} \\ $$$$\Rightarrow{u}^{\mathrm{2}} =\frac{\mathrm{4}}{{u}}\:\Rightarrow{u}^{\mathrm{3}} =\mathrm{4} \\ $$$$\Rightarrow{u}=\sqrt[{\mathrm{3}}]{\mathrm{4}}\:\Rightarrow\mathrm{log}\:_{{x}} \left({y}\right)=\sqrt[{\mathrm{3}}]{\mathrm{4}} \\ $$$$\Rightarrow{y}={x}^{\sqrt[{\mathrm{3}}]{\mathrm{4}}} \: \\ $$$$\Rightarrow\mathrm{log}_{{x}^{\sqrt[{\mathrm{3}}]{\mathrm{4}}} } \:\left({x}\right)=\mathrm{log}_{{x}} \left(\mathrm{2}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{4}}}\:\mathrm{log}_{{x}} \left({x}\right)=\mathrm{log}_{{x}} \left(\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{2}^{\sqrt[{\mathrm{3}}]{\mathrm{4}}} \:={x}\:\:\: \\ $$