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Question Number 162088 by mathlove last updated on 26/Dec/21

x^2 =2^x solve for x=?

$${x}^{\mathrm{2}} =\mathrm{2}^{{x}} \\ $$$${solve}\:\:\:{for}\:\:\:\:{x}=? \\ $$

Answered by Ar Brandon last updated on 26/Dec/21

x^2 =2^x 2lnx=xln2 (1/x)lnx=(1/2)ln2 −lnx∙e^(−lnx) =−((ln2)/2) W(−lnx)=−((ln2)/2) −lnx=W_0 (−((ln2)/2)) x=e^(−W_0 (−((ln2)/2)))

$${x}^{\mathrm{2}} =\mathrm{2}^{{x}} \\ $$$$\mathrm{2ln}{x}={x}\mathrm{ln2} \\ $$$$\frac{\mathrm{1}}{{x}}\mathrm{ln}{x}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln2} \\ $$$$−\mathrm{ln}{x}\centerdot{e}^{−\mathrm{ln}{x}} =−\frac{\mathrm{ln2}}{\mathrm{2}} \\ $$$$\mathrm{W}\left(−\mathrm{ln}{x}\right)=−\frac{\mathrm{ln2}}{\mathrm{2}} \\ $$$$−\mathrm{ln}{x}={W}_{\mathrm{0}} \left(−\frac{\mathrm{ln2}}{\mathrm{2}}\right) \\ $$$${x}={e}^{−{W}_{\mathrm{0}} \left(−\frac{\mathrm{ln2}}{\mathrm{2}}\right)} \\ $$

Commented by mathlove last updated on 26/Dec/21

The numerical answer is

$$ \\ $$The numerical answer is

Commented by mr W last updated on 26/Dec/21

W(−((ln 2)/2)) has two values W(−((ln 2)/2))= { ((−1.3862944)),((−0.6931472)) :} therefore there are two solutions x= { ((e^(1.3862944) =4)),((e^(0.6931472) =2)) :}

$${W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right)\:{has}\:{two}\:{values} \\ $$$${W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right)=\begin{cases}{−\mathrm{1}.\mathrm{3862944}}\\{−\mathrm{0}.\mathrm{6931472}}\end{cases} \\ $$$${therefore}\:{there}\:{are}\:{two}\:{solutions} \\ $$$${x}=\begin{cases}{{e}^{\mathrm{1}.\mathrm{3862944}} =\mathrm{4}}\\{{e}^{\mathrm{0}.\mathrm{6931472}} =\mathrm{2}}\end{cases} \\ $$

Commented by Ar Brandon last updated on 26/Dec/21

Thank you, Sir.

$$\mathrm{Thank}\:\mathrm{you},\:\mathrm{Sir}. \\ $$

Commented by Ar Brandon last updated on 26/Dec/21

How to get the 2 values? I got only 0.69... using the series expansion.

$$\mathrm{How}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{2}\:\mathrm{values}?\:\mathrm{I}\:\mathrm{got}\:\mathrm{only}\:\mathrm{0}.\mathrm{69}... \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{series}\:\mathrm{expansion}. \\ $$

Commented by mr W last updated on 27/Dec/21

W(−((ln 2)/2)) is the root(s) of xe^x =−((ln 2)/2) there are two roots. i got the numeric values using Grapher:

$${W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right)\:{is}\:{the}\:{root}\left({s}\right)\:{of} \\ $$$${xe}^{{x}} =−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}} \\ $$$${there}\:{are}\:{two}\:{roots}.\:{i}\:{got}\:{the} \\ $$$${numeric}\:{values}\:{using}\:{Grapher}: \\ $$

Commented by mathlove last updated on 27/Dec/21

thanks sir

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

Commented by mr W last updated on 27/Dec/21

Answered by Rasheed.Sindhi last updated on 27/Dec/21

Comparing exponents on both sides and comparing bases on both sides it′s obvious that x=2. x⇆2 So x=2

$${Comparing}\:{exponents}\:{on}\:{both}\:{sides} \\ $$$${and}\:{comparing}\:{bases}\:{on}\:{both}\:{sides} \\ $$$${it}'{s}\:{obvious}\:{that}\:{x}=\mathrm{2}. \\ $$$$\:{x}\leftrightarrows\mathrm{2} \\ $$$${So}\:{x}=\mathrm{2} \\ $$

Commented by mr W last updated on 26/Dec/21

2^2 =2^2 ⇒x=2 and 4^2 =2^4 ⇒x=4

$$\mathrm{2}^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} \:\Rightarrow{x}=\mathrm{2} \\ $$$${and} \\ $$$$\mathrm{4}^{\mathrm{2}} =\mathrm{2}^{\mathrm{4}} \:\Rightarrow{x}=\mathrm{4} \\ $$

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