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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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