solve-for-x-by-using-lambert-function-x-2-16-x-

Question Number 183756 by Michaelfaraday last updated on 29/Dec/22

$${solve}\:{for}\:{x}\:{by}\:{using}\:{lambert}\:{function} \\ $$$${x}^{\mathrm{2}} =\mathrm{16}^{{x}} \\ $$

Answered by aleks041103 last updated on 30/Dec/22

2lnx=xln16 ⇒lnx=xln4 (1/x)ln((1/x))=−ln(4) ⇒ye^y =e^(−ln(4)) =(1/4)⇒(1/x)=W((1/4)) ⇒x=(1/(W(1/4)))

$$\mathrm{2}{lnx}={xln}\mathrm{16} \\ $$$$\Rightarrow{lnx}={xln}\mathrm{4} \\ $$$$\frac{\mathrm{1}}{{x}}{ln}\left(\frac{\mathrm{1}}{{x}}\right)=−{ln}\left(\mathrm{4}\right) \\ $$$$\Rightarrow{ye}^{{y}} ={e}^{−{ln}\left(\mathrm{4}\right)} =\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow\frac{\mathrm{1}}{{x}}={W}\left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{{W}\left(\mathrm{1}/\mathrm{4}\right)} \\ $$

Commented by Michaelfaraday last updated on 30/Dec/22

sir,are you sure about your solution because when i solve it, i arrive at x=((−W(((−1)/2)In16))/((1/2)In16))

$${sir},{are}\:{you}\:{sure}\:{about}\:{your}\:{solution} \\ $$$${because}\:{when}\:{i}\:{solve}\:{it},\:{i}\:{arrive}\:{at} \\ $$$${x}=\frac{−{W}\left(\frac{−\mathrm{1}}{\mathrm{2}}{In}\mathrm{16}\right)}{\frac{\mathrm{1}}{\mathrm{2}}{In}\mathrm{16}} \\ $$

Commented by mr W last updated on 30/Dec/22

you are wrong, too. x=−((W(((ln 16)/2)))/((ln 16)/2))=−((W(ln 4))/(ln 4))=−(1/2)

$${you}\:{are}\:{wrong},\:{too}. \\ $$$${x}=−\frac{{W}\left(\frac{\mathrm{ln}\:\mathrm{16}}{\mathrm{2}}\right)}{\frac{\mathrm{ln}\:\mathrm{16}}{\mathrm{2}}}=−\frac{{W}\left(\mathrm{ln}\:\mathrm{4}\right)}{\mathrm{ln}\:\mathrm{4}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by Michaelfaraday last updated on 30/Dec/22

$${but}\:{i}\:{got}\:{it}\:{sir}\:{but}\:{i}\:{just}\:{dont}\:{simplify}\:{it} \\ $$$${to}\:{lowest}\:{term}. \\ $$

Commented by mr W last updated on 30/Dec/22

i didn′t mean that! check again! you got: x=((−W(((−1)/2)In16))/((1/2)In16)) this is wrong.

$${i}\:{didn}'{t}\:{mean}\:{that}!\:{check}\:{again}!\: \\ $$$${you}\:{got}: \\ $$$${x}=\frac{−{W}\left(\frac{−\mathrm{1}}{\mathrm{2}}{In}\mathrm{16}\right)}{\frac{\mathrm{1}}{\mathrm{2}}{In}\mathrm{16}} \\ $$$${this}\:{is}\:{wrong}. \\ $$

Commented by Michaelfaraday last updated on 30/Dec/22

$${oh}\:{have}\:{seen}\:{it}\:{but}\:{i}\:{made}\:{a}\:{typo}\:{in}\:{answer} \\ $$$${thst}\:{is}\:{why},{thanks} \\ $$

Answered by MJS_new last updated on 30/Dec/22

x^2 =16^x let x=−e^(−t) e^(−2t) =16^(−e^(−t) ) −2t=−e^(−t) 4ln 2 e^t t=2ln 2 obviously t=ln 2 is the solution ⇒ x=−(1/2)

$${x}^{\mathrm{2}} =\mathrm{16}^{{x}} \\ $$$$\mathrm{let}\:{x}=−\mathrm{e}^{−{t}} \\ $$$$\mathrm{e}^{−\mathrm{2}{t}} =\mathrm{16}^{−\mathrm{e}^{−{t}} } \\ $$$$−\mathrm{2}{t}=−\mathrm{e}^{−{t}} \mathrm{4ln}\:\mathrm{2} \\ $$$$\mathrm{e}^{{t}} {t}=\mathrm{2ln}\:\mathrm{2} \\ $$$$\mathrm{obviously}\:{t}=\mathrm{ln}\:\mathrm{2}\:\mathrm{is}\:\mathrm{the}\:\mathrm{solution} \\ $$$$\Rightarrow\:{x}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by Michaelfaraday last updated on 30/Dec/22

$${thanks}\:{sir} \\ $$

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