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)
$$ \\ $$$${ln}\left({x}^{{ln}\left(\mathrm{2}{x}\right)} \right)={ln}\left(\mathrm{5}\right) \\ $$$${ln}\left(\mathrm{2}{x}\right){ln}\left({x}\right)={ln}\left(\mathrm{5}\right) \\ $$$$\left[{ln}\left(\mathrm{2}\right)+{ln}\left({x}\right)\right]{ln}\left({x}\right)={ln}\left(\mathrm{5}\right) \\ $$$${ln}^{\mathrm{2}} \left({x}\right)+{ln}\left(\mathrm{2}\right){ln}\left({x}\right)−{ln}\left(\mathrm{5}\right)=\mathrm{0} \\ $$$$ \\ $$$$\therefore\:{ln}\left({x}\right)=\frac{−{ln}\left(\mathrm{2}\right)\pm\sqrt{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\mathrm{4}×\mathrm{1}×−{ln}\left(\mathrm{5}\right)}}{\mathrm{2}×\mathrm{1}} \\ $$$$ \\ $$$${ln}\left({x}\right)=\frac{−{ln}\left(\mathrm{2}\right)\pm\sqrt{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\mathrm{4}{ln}\left(\mathrm{5}\right)}}{\mathrm{2}} \\ $$$$\Leftrightarrow\:{x}={e}^{\frac{−{ln}\left(\mathrm{2}\right)\pm\sqrt{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\mathrm{4}{ln}\left(\mathrm{5}\right)}}{\mathrm{2}}} \\ $$
Answered by TonyCWX last updated on 07/Jan/26
![A logarithm without base is assumed as log_(10) . x^(log(2x)) = 5 log(2x)log(x) = log(5) log(x)[log(2)+log(x)] = log(5) log(2)log(x)+log^2 (x) = log(5) log^2 (x) + log(2)log(x)−log(5) = 0 log(x) = ((−log(2)±(√(log^2 (2)+4log(5))))/2) log(x) = ((−log(2)+(√(log^2 (2)+4log(5))))/2) x = 10^((−log(2)+(√(log^2 (2)+4log(5))))/2) = 5](https://www.tinkutara.com/question/Q227229.png)
$$\mathrm{A}\:\mathrm{logarithm}\:\mathrm{without}\:\mathrm{base}\:\mathrm{is}\:\mathrm{assumed}\:\mathrm{as}\:\mathrm{log}_{\mathrm{10}} . \\ $$$$\mathrm{x}^{\mathrm{log}\left(\mathrm{2x}\right)} \:=\:\mathrm{5} \\ $$$$\mathrm{log}\left(\mathrm{2x}\right)\mathrm{log}\left(\mathrm{x}\right)\:=\:\mathrm{log}\left(\mathrm{5}\right) \\ $$$$\mathrm{log}\left(\mathrm{x}\right)\left[\mathrm{log}\left(\mathrm{2}\right)+\mathrm{log}\left(\mathrm{x}\right)\right]\:=\:\mathrm{log}\left(\mathrm{5}\right) \\ $$$$\mathrm{log}\left(\mathrm{2}\right)\mathrm{log}\left(\mathrm{x}\right)+\mathrm{log}^{\mathrm{2}} \left(\mathrm{x}\right)\:=\:\mathrm{log}\left(\mathrm{5}\right) \\ $$$$\mathrm{log}^{\mathrm{2}} \left(\mathrm{x}\right)\:+\:\mathrm{log}\left(\mathrm{2}\right)\mathrm{log}\left(\mathrm{x}\right)−\mathrm{log}\left(\mathrm{5}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{log}\left(\mathrm{x}\right)\:=\:\frac{−\mathrm{log}\left(\mathrm{2}\right)\pm\sqrt{\mathrm{log}^{\mathrm{2}} \left(\mathrm{2}\right)+\mathrm{4log}\left(\mathrm{5}\right)}}{\mathrm{2}} \\ $$$$\mathrm{log}\left(\mathrm{x}\right)\:=\:\frac{−\mathrm{log}\left(\mathrm{2}\right)+\sqrt{\mathrm{log}^{\mathrm{2}} \left(\mathrm{2}\right)+\mathrm{4log}\left(\mathrm{5}\right)}}{\mathrm{2}} \\ $$$$\mathrm{x}\:=\:\mathrm{10}^{\frac{−\mathrm{log}\left(\mathrm{2}\right)+\sqrt{\mathrm{log}^{\mathrm{2}} \left(\mathrm{2}\right)+\mathrm{4log}\left(\mathrm{5}\right)}}{\mathrm{2}}} \:=\:\mathrm{5} \\ $$