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Question Number 93546 by allizzwell23 last updated on 13/May/20

$$\mathrm{Which}\mathrm{app}\mathrm{is}\mathrm{the}\mathrm{best}\mathrm{to}\mathrm{evaluate}$$$$\frac{\mathrm{W}\left(2\ln 2\right)}{\ln 2}$$

Commented by IbrahimGR376 last updated on 13/May/20

$$\mathrm{mathematica}\mathrm{app}$$$$\mathrm{GR}376$$

Commented by prakash jain last updated on 13/May/20

https://www.wolframalpha.com

Commented by allizzwell23 last updated on 13/May/20

$$\mathrm{i}\mathrm{dont}\mathrm{know}\mathrm{how}\mathrm{to}\mathrm{input}\mathrm{correctly}$$

Commented by prakash jain last updated on 13/May/20

in wolfram alpha. You can simply write plain text W(2ln2)/ln2

Commented by allizzwell23 last updated on 13/May/20

$$\mathrm{yes}\mathrm{sir}.\mathrm{can}\mathrm{you}\mathrm{show}\mathrm{me}\mathrm{how}\mathrm{to}\mathrm{use}\mathrm{it}$$

Commented by prakash jain last updated on 13/May/20

Wolfram alpha is just plain text input

Commented by prakash jain last updated on 13/May/20

Wolfram will give alternate implementations of W. It will hive u options to select. for specific interpretations you can lambertw(2ln2)/ln2. or eta(5) or theta(5)

Commented by allizzwell23 last updated on 13/May/20

$$\mathrm{thanks}\mathrm{sir}$$

Commented by mr W last updated on 13/May/20

$$becareful:$$$$forx<0,wolframalphagivesonly$$$$onevalueofW\left(x\right).forexample:$$$$W(-0.1)=-0.111833or-3.577152.$$$$butwolframalphagivesonly$$$$W(-0.1)=-0.111833$$

Commented by mr W last updated on 13/May/20

$$butifyoujustwanttoget\frac{W\left(2\ln 2\right)}{\ln 2},$$$$you{don}^{\prime }tneedanyapp.itis1.$$$$i.e.\frac{W\left(2\ln 2\right)}{\ln 2}=1.$$$$how?$$$$ac.todefinition:$$$$W\left(2\ln 2\right)e^{W\left(2\ln 2\right)}=2\ln 2=\left(\ln 2\right)e^{\left(\ln 2\right)}$$$$\Rightarrow W\left(2\ln 2\right)=\ln 2$$$$\Rightarrow \frac{W\left(2\ln 2\right)}{\ln 2}=1$$

Commented by prakash jain last updated on 13/May/20

$${ye}^y=x$$$$x⩾0\mathrm{we}\mathrm{get}\mathrm{one}\mathrm{value}\mathrm{for}y\mathrm{which}$$$$\mathrm{is}\mathrm{given}\mathrm{by}{W}_0\left(x\right)$$$$-\frac{1}{e}⩽x<0$$$$\mathrm{We}\mathrm{get}\mathrm{two}\mathrm{values}\mathrm{of}x\mathrm{which}$$$$\mathrm{are}\mathrm{given}{W}_0\left(x\right)\mathrm{and}{W}_{-1}\left(x\right)$$$$\mathrm{Wolform}\mathrm{will}\mathrm{correctly}\mathrm{give}$$$$\mathrm{two}\mathrm{results}\mathrm{when}\mathrm{you}\mathrm{ask}\mathrm{it}$$$$\mathrm{to}\mathrm{solve}{xe}^x=-.1$$$$\mathrm{but}W\left(x\right)\mathrm{it}\mathrm{will}\mathrm{interpret}\mathrm{as}{W}_0\left(x\right)$$$$\mathrm{which}\mathrm{is}\mathrm{a}\mathrm{single}\mathrm{valued}\mathrm{function}$$$$\mathrm{for}x=-.1$$

Commented by allizzwell23 last updated on 13/May/20

$$\mathrm{thanks}$$

Commented by mr W last updated on 13/May/20

$$thanksfortheexplanationsir!$$

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