Question

Solve for x:
e^x^2 - 8 x^2=4

Answer 1

why?

this can't really be done with standard methods

Answer 2

sure it can..

Answer 3

differentiating both sides may work

Answer 4

what is the answer? seems it is in natural log form

Answer 5

try this way
e^x^2 - 8 x^2=4
e^x^2 = 8 x^2+ 4
taking logs:
x^2 = ln (8x^2 + 4)

hmm not sure to be honest...

Answer 6

sqrt(ln 8) ?

Answer 7

you might try a graphical method
draw graphs of x^2 and the log function on same axes

Answer 8

been a while since i studied differentiation, but here goes

differentiating both sides we get
(2x)*e^x^2 -16x =0
if x is not = 0

2*e^x^2 =16
or e^x^2 =8

taking natural log both sides

x^2 =ln (8)
x=sqrt{ln(8)}

whatever this value is......
I may be wrong in my calculations since its been a few years i studied calculus.

Answer 9

using graphs i get solution set (-1.86 , 1.86)

Answer 10

differentiation doesn't tell us anything. finding x for f'(x) = 0 is the location where the slope = 0

Answer 11

yes

Answer 12

hmm .....forgot

Answer 13

the answers i gave check out but there might be an algebraic method to do this but can't think of one

Answer 14

there must be an algebraic method,y'all!

Answer 15

There is NOT an algebraic method. Even if you simplify it with substitute, \(e^{y} - 8y = 4\), there is NOT algebraic solution. Take a good, hard look at Lambert's W-Function and you will see the similarity and the problem.

\(x = \pm\sqrt{ln(8)}\) is the location of the symmetrical absolute minima. The solutions must be outside this range. \(\sqrt{ln(8)}\) might be a good place to start looking, but you should find THEM with slightly greater magnitude.

\(\sqrt{ln(8)} = 1.4420268866\) This gives @cwrw238 's graphical solution of +/- 1.86 good credibility.

Newton's Method (or just about any other numerical method) MAY find them quickly in this situation. We have to find only one and the other is trivial, by symmetry.

Actually, Newton's Method fails miserably when starting at \(x = \sqrt{ln(8)}\). You can get there when starting with x = 1.5, but we already have @cwrw238 's 1.86, so let's start there.

Indeed, it converges VERY quickly.

\(x_{0} = 1.86\)
\(x_{1} = 1.85856070410157\)
\(x_{2} = 1.85855496698982\)
\(x_{3} = 1.85855496689909\)

So, there they are, x = +1.85855496689909 or x = -1.85855496689909

You CANNOT do this with simple algebraic manipulation. There is NOT a nice, closed-form expression, unless you consider the Lambert W-Function amongst the tools that constitute "simple" or "closed-form".

Answer 16

Note: When I say there ISN'T a way to do it, this does not mean we're just not clever enough. It does not mean that if we keep trying, we might get it. It can't be done!! That is what it means.

Answer 17

The base case is this, \(x = e^{x}\). Can't do it. Try as you might, you will not succeed.
Lambert worked with this: \(W(x) = xe^{x}\). Very similar.

It can be disappointing to a sharp algebra student to discover that such simple expressions do NOT have lovely algebraic solutions or manipulations that help us solve them. Nevertheless, it is the way it is.

Answer 18

yep tkhunny is right

Answer 19

thanks people!

Answer 20

yw