solve-for-x-the-equation-log-x-e-2x-eln-x-e-

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

s o l v e f o r x t h e e q u a t i o n

{l o g}_{x} e^{2 x} = e l n x - e

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

\frac{2 x}{l n x} = e l n x - e \Rightarrow e {(l n x)}^{2} - e l n x - 2 x = 0

s e t y = l n x \Rightarrow x = e^{y}

\Rightarrow {e y}^{2} - e y - 2 e^{y} = 0 \Rightarrow y^{2} - y = 2 e^{y - 1}

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

m a y b e, w e c a n f i n d t h e a n s w e r s b y d r o w

y^{2} - y a n d 2 e^{y - 1} .

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

r e a l l y t h a t w i l l b e u n e a s y . . t h a n k s a n y w a y

Commented by MJS last updated on 17/Sep/19

we can only approximate

eln x (\ln x - 1) - 2 x = 0

\Rightarrow

x \approx . 690331

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

t h a n k s s o m u c h, b u t h o w d i d y o u g e t t h a t a p p r o x i m a t e

v a l u e, i m e a n t h e s t e p s t o o b t a i n s u c h a p p r o x i m a t e

Commented by MJS last updated on 18/Sep/19

first, plot it as a function

y = eln x (\ln x - 1) - 2 x

or calculate some values

we know x > 0 (because \ln x is not defined

for x ⩽ 0)

put x = \frac{1}{e} \Rightarrow \ln x = - 1 \Rightarrow y = 2 e - \frac{2}{e} > 0

put x = 1 \Rightarrow \ln x = 0 \Rightarrow y = - 2 < 0

\Rightarrow \frac{1}{e} \approx . 367879 < x < 1

now try x = {. 4, . 5, . 6, . 7, . 8, . 9}

x = . 6 \Rightarrow y = . 897 \dots

x = . 7 \Rightarrow y = - . 0846 \dots

\Rightarrow . 6 < x < . 7 but {it}^{'} s closer to . 7

\Rightarrow try x = {. 69, . 68, . 67, \dots, . 61}

\Rightarrow . 69 < x < . 70 but {it}^{'} s closer to . 69

\Rightarrow try x = {. 691, . 692, \dots, . 699}

x = . 691 \Rightarrow y = - . 00592 \dots

\Rightarrow try x = {. 6901, . 6902, \dots

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

t h a n k y o u s o m u c h