Đáp án: $x\in\{-e^{-W\left(\dfrac{\ln2}{2}\right)},e^{-W\left(-\dfrac{\ln 2}{2}\right)}\}$
Giải thích các bước giải:
Bạn tìm hiểu thêm về hàm Lambert nhé.
Ký hiệu hàm Lambert: $W\left(x\right)=f^{-1}\left(x\right)$ là hàm ngược của hàm $f\left(x\right)$ với $f\left(x\right)=xe^x$
$\to W\left(x\right)=f^{-1}\left(xe^x\right)=x$
Ta có:
$x^2=2^x$
$\to \ln\left(x^2\right)=\ln 2^x$
$\to 2\ln|x|=x\ln2$
Trường hợp $x>0$
$\to 2\ln x=x\ln 2$
$\to \dfrac{\ln x}{x}=\dfrac{\ln 2}{2}$
$\to \dfrac{\ln x}{e^{\ln x}}=\dfrac{\ln 2}{2}$
$\to \ln xe^{-\ln x}=\dfrac{\ln 2}{2}$
$\to \left(-\ln x\right)e^{-\ln x}=-\dfrac{\ln 2}{2}$
$\to W\left(\left(-\ln x\right)e^{-\ln x}\right)=W\left(-\dfrac{\ln 2}{2}\right)$
$\to -\ln x=W\left(-\dfrac{\ln 2}{2}\right)$
$\to \ln x=-W\left(-\dfrac{\ln 2}{2}\right)$
$\to x=e^{-W\left(-\dfrac{\ln 2}{2}\right)}$
Trường hợp $x<0$
$\to 2\ln\left(-x\right)=x\ln2$
$\to \dfrac{\ln\left(-x\right)}{x}=\dfrac{\ln2}{2}$
$\to \dfrac{\ln\left(-x\right)}{-x}=-\dfrac{\ln2}{2}$
$\to \dfrac{\ln\left(-x\right)}{e^{\ln\left(-x\right)}}=-\dfrac{\ln2}{2}$
$\to \ln\left(-x\right)e^{-\ln\left(-x\right)}=-\dfrac{\ln2}{2}$
$\to \left(-\ln\left(-x\right)\right)e^{-\ln\left(-x\right)}=\dfrac{\ln2}{2}$
$\to W\left(\left(-\ln\left(-x\right)\right)e^{-\ln\left(-x\right)}\right)=W\left(\dfrac{\ln2}{2}\right)$
$\to -\ln\left(-x\right)=W\left(\dfrac{\ln2}{2}\right)$
$\to \ln\left(-x\right)=-W\left(\dfrac{\ln2}{2}\right)$
$\to -x=e^{-W\left(\dfrac{\ln2}{2}\right)}$
$\to x=-e^{-W\left(\dfrac{\ln2}{2}\right)}$
