Question Number 151403 by 7770 last updated on 20/Aug/21
$$\:{x}^{\mathrm{3}} =\mathrm{3}^{{x}} \\ $$$${how}\:{do}\:{i}\:{pls}\:{use}\:{lambert}\:{to}\: \\ $$$$\:{find}\:{x}=\mathrm{3} \\ $$
Commented by mr W last updated on 20/Aug/21
$${x}=\mathrm{3}^{\frac{{x}}{\mathrm{3}}} \\ $$$${x}={e}^{\frac{{x}}{\mathrm{3}}\mathrm{ln}\:\mathrm{3}} \\ $$$${xe}^{−\frac{{x}}{\mathrm{3}}\mathrm{ln}\:\mathrm{3}} =\mathrm{1} \\ $$$$\left(−\frac{{x}}{\mathrm{3}}\mathrm{ln}\:\mathrm{3}\right){e}^{−\frac{{x}}{\mathrm{3}}\mathrm{ln}\:\mathrm{3}} =−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{3}} \\ $$$$−\frac{{x}}{\mathrm{3}}\mathrm{ln}\:\mathrm{3}={W}\left(−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{3}}\right) \\ $$$$\Rightarrow{x}=−\frac{\mathrm{3}}{\mathrm{ln}\:\mathrm{3}}{W}\left(−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{3}}\right) \\ $$$$=\begin{cases}{−\frac{\mathrm{3}}{\mathrm{ln}\:\mathrm{3}}×\left(−\mathrm{1}.\mathrm{098612289}\right)=\mathrm{3}}\\{−\frac{\mathrm{3}}{\mathrm{ln}\:\mathrm{3}}×\left(−\mathrm{0}.\mathrm{907473042}\right)=\mathrm{2}.\mathrm{478053}}\end{cases} \\ $$
Commented by 7770 last updated on 20/Aug/21
$${thank}\:{you}\:{sir}\:{but}\:{how}\:{did}\:{u}\:{get}\:\left(−\mathrm{1}.\mathrm{098612289}\right)\:{please} \\ $$
Commented by mr W last updated on 21/Aug/21
$${use}\:{any}\:{method}\:{for}\:{solving}\:{equations} \\ $$$${numerically}.\:{the}\:{equation}\:{is} \\ $$$${xe}^{{x}} ={a} \\ $$
Commented by 7770 last updated on 21/Aug/21
$${ok}\:{sir} \\ $$$${thanks} \\ $$
Commented by otchereabdullai@gmail.com last updated on 22/Aug/21
$$\mathrm{thank}\:\mathrm{u} \\ $$