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

solve for x the equation log_x e^(2x) = eln x −e

$${solve}\:{for}\:{x}\:{the}\:{equation} \\ $$$$\:\:{log}_{{x}} {e}^{\mathrm{2}{x}} \:=\:{eln}\:{x}\:−{e} \\ $$

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

((2x)/(lnx))=elnx−e⇒e(lnx)^2 −elnx−2x=0 set y=lnx⇒x=e^y ⇒ey^2 −ey−2e^y =0⇒y^2 −y=2e^(y−1)

$$\frac{\mathrm{2}{x}}{{lnx}}={elnx}−{e}\Rightarrow{e}\left({lnx}\right)^{\mathrm{2}} −{elnx}−\mathrm{2}{x}=\mathrm{0} \\ $$$${set}\:{y}={lnx}\Rightarrow{x}={e}^{{y}} \\ $$$$\Rightarrow{ey}^{\mathrm{2}} −{ey}−\mathrm{2}{e}^{{y}} =\mathrm{0}\Rightarrow{y}^{\mathrm{2}} −{y}=\mathrm{2}{e}^{{y}−\mathrm{1}} \\ $$

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

maybe,we can find the answers by drow y^2 −y and 2e^(y−1) .

$${maybe},{we}\:{can}\:{find}\:{the}\:{answers}\:{by}\:{drow} \\ $$$${y}^{\mathrm{2}} −{y}\:{and}\:\mathrm{2}{e}^{{y}−\mathrm{1}} . \\ $$

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

really that will be uneasy..thanks anyway

$${really}\:{that}\:{will}\:{be}\:{uneasy}..{thanks}\:{anyway} \\ $$

Commented by MJS last updated on 17/Sep/19

we can only approximate eln x (ln x −1)−2x=0 ⇒ x≈.690331

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

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

thanks so much, but how did you get that approximate value, i mean the steps to obtain such approximate

$${thanks}\:{so}\:{much},\:{but}\:{how}\:{did}\:{you}\:{get}\:{that}\:{approximate}\: \\ $$$${value},\:{i}\:{mean}\:{the}\:{steps}\:{to}\:{obtain}\:{such}\:{approximate} \\ $$

Commented by MJS last updated on 18/Sep/19

first, plot it as a function y=eln x (ln x −1)−2x or calculate some values we know x>0 (because ln x is not defined for x≤0) put x=(1/e) ⇒ ln x =−1 ⇒ y=2e−(2/e)>0 put x=1 ⇒ ln x =0 ⇒ y=−2<0 ⇒ (1/e)≈.367879<x<1 now try x={.4, .5, .6, .7, .8, .9} x=.6 ⇒ y=.897... x=.7 ⇒ y=−.0846... ⇒ .6<x<.7 but it′s closer to .7 ⇒ try x={.69, .68, .67, ..., .61} ⇒ .69<x<.70 but it′s closer to .69 ⇒ try x={.691, .692, ..., .699} x=.691 ⇒ y=−.00592... ⇒ try x={.6901, .6902, ...

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

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

thank you so much

$${thank}\:{you}\:{so}\:{much} \\ $$

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