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Question Number 68898 by Rio Michael last updated on 16/Sep/19

$$solveforxtheequation$$$${log}_x{e}^{2x}=elnx-e$$

Commented by kaivan.ahmadi last updated on 16/Sep/19

$$\frac{2x}{lnx}=elnx-e\Rightarrow e{\left(lnx\right)}^2-elnx-2x=0$$$$sety=lnx\Rightarrow x=e^y$$$$\Rightarrow {ey}^2-ey-2e^y=0\Rightarrow y^2-y=2e^{y-1}$$

Commented by kaivan.ahmadi last updated on 16/Sep/19

$$maybe,wecanfindtheanswersbydrow$$$$y^2-yand2e^{y-1}.$$

Commented by Rio Michael last updated on 17/Sep/19

$$reallythatwillbeuneasy..thanksanyway$$

Commented by MJS last updated on 17/Sep/19

$$\mathrm{we}\mathrm{can}\mathrm{only}\mathrm{approximate}$$$$\mathrm{eln}x(\ln x-1)-2x=0$$$$\Rightarrow$$$$x\approx .690331$$

Commented by Rio Michael last updated on 17/Sep/19

$$thankssomuch,buthowdidyougetthatapproximate$$$$value,imeanthestepstoobtainsuchapproximate$$

Commented by MJS last updated on 18/Sep/19

$$\mathrm{first},\mathrm{plot}\mathrm{it}\mathrm{as}\mathrm{a}\mathrm{function}$$$$y=\mathrm{eln}x(\ln x-1)-2x$$$$\mathrm{or}\mathrm{calculate}\mathrm{some}\mathrm{values}$$$$\mathrm{we}\mathrm{know}x>0(\mathrm{because}\ln x\mathrm{is}\mathrm{not}\mathrm{defined}$$$$\mathrm{for}x⩽0)$$$$\mathrm{put}x=\frac{1}{\mathrm{e}}\Rightarrow \ln x=-1\Rightarrow y=2\mathrm{e}-\frac{2}{\mathrm{e}}>0$$$$\mathrm{put}x=1\Rightarrow \ln x=0\Rightarrow y=-2<0$$$$\Rightarrow \frac{1}{\mathrm{e}}\approx .367879<x<1$$$$\mathrm{now}\mathrm{try}x=\{.4,.5,.6,.7,.8,.9\}$$$$x=.6\Rightarrow y=.897...$$$$x=.7\Rightarrow y=-.0846...$$$$\Rightarrow .6<x<.7\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{closer}\mathrm{to}.7$$$$\Rightarrow \mathrm{try}x=\{.69,.68,.67,...,.61\}$$$$\Rightarrow .69<x<.70\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{closer}\mathrm{to}.69$$$$\Rightarrow \mathrm{try}x=\{.691,.692,...,.699\}$$$$x=.691\Rightarrow y=-.00592...$$$$\Rightarrow \mathrm{try}x=\{.6901,.6902,...$$

Commented by Rio Michael last updated on 18/Sep/19

$$thankyousomuch$$

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