Question Number 227192 by gregori last updated on 05/Jan/26
$$\:\:\:\:{x}^{\mathrm{log}\:\mathrm{2}{x}} \:=\:\mathrm{5}\:\Rightarrow{x}=\:? \\ $$
Answered by Kassista last updated on 05/Jan/26
![ln(x^(ln(2x)) )=ln(5) ln(2x)ln(x)=ln(5) [ln(2)+ln(x)]ln(x)=ln(5) ln^2 (x)+ln(2)ln(x)−ln(5)=0 ∴ ln(x)=((−ln(2)±(√(ln^2 (2)−4×1×−ln(5))))/(2×1)) ln(x)=((−ln(2)±(√(ln^2 (2)+4ln(5))))/2) ⇔ x=e^((−ln(2)±(√(ln^2 (2)+4ln(5))))/2)](https://www.tinkutara.com/question/Q227193.png)
$$ \\ $$$${ln}\left({x}^{{ln}\left(\mathrm{2}{x}\right)} \right)={ln}\left(\mathrm{5}\right) \\ $$$${ln}\left(\mathrm{2}{x}\right){ln}\left({x}\right)={ln}\left(\mathrm{5}\right) \\ $$$$\left[{ln}\left(\mathrm{2}\right)+{ln}\left({x}\right)\right]{ln}\left({x}\right)={ln}\left(\mathrm{5}\right) \\ $$$${ln}^{\mathrm{2}} \left({x}\right)+{ln}\left(\mathrm{2}\right){ln}\left({x}\right)−{ln}\left(\mathrm{5}\right)=\mathrm{0} \\ $$$$ \\ $$$$\therefore\:{ln}\left({x}\right)=\frac{−{ln}\left(\mathrm{2}\right)\pm\sqrt{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\mathrm{4}×\mathrm{1}×−{ln}\left(\mathrm{5}\right)}}{\mathrm{2}×\mathrm{1}} \\ $$$$ \\ $$$${ln}\left({x}\right)=\frac{−{ln}\left(\mathrm{2}\right)\pm\sqrt{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\mathrm{4}{ln}\left(\mathrm{5}\right)}}{\mathrm{2}} \\ $$$$\Leftrightarrow\:{x}={e}^{\frac{−{ln}\left(\mathrm{2}\right)\pm\sqrt{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\mathrm{4}{ln}\left(\mathrm{5}\right)}}{\mathrm{2}}} \\ $$