# Solving an exponential function

1. Dec 12, 2016

### chwala

1. The problem statement, all variables and given/known data

Solve the equation
$e^{2x}+2=e^{3x-4}$

2. Relevant equations

3. The attempt at a solution
I know by using Newton-Raphson method the problem can be solved, i however tried solving it as follows
$e^{3x-4}-e^{2x}-2=0, e^{3x}-e^{2x}.e^{4}-2e^{4}=0, p^3-p^2.e^4-2e^4=0, →p^3-54.6p^2-109.2=0$
where
$p=e^x$,
am i correct, will i get solution this way

Last edited: Dec 12, 2016
2. Dec 12, 2016

### ehild

Yes, you can do it.

3. Dec 13, 2016

### chwala

i will post solution soon, i dont have calculator now but i know solution will be of the form $x=ln p$

4. Dec 13, 2016

### chwala

the solution is thus $p=54.64→x=ln 54.64, ⇒x=4.00$

5. Dec 13, 2016

### ehild

Yes, very close to it.

6. Dec 13, 2016

### chwala

yes, $x=4.00$ is an approximate solution to the problem.

7. Dec 13, 2016

### chwala

Might there be other approximate solutions to the problem, i ignored the negative values of $p=-0.0183+1.41i, -0.0183-1.41i$

8. Dec 13, 2016

### Ray Vickson

You will get an approximation. To get better accuracy, you need a more accurate value for $e^4$. In principle you can write down an exact solution, using exact formulas for solutions of a cubic equation and retaining $e^4$ in symbolic form.

9. Dec 13, 2016

### Ray Vickson

You can see that $x=4$ cannot be the solution, but is very close to a solution. For $x = 4$ the left-hand-side is $2+e^{8\times 4} = 2 + e^8$ while the right-hand-side is $e^{3 \times 4 - 4} = e^8$. Since $e^8 \doteq 2980.958$ the two sides of the equation are close in terms of relative magnitude (that is, % difference), although they still differ by 2 in absolute terms.

You can also see how to get a good approximation quickly: set $x = 4 + y$ with $y$ small. The equation becomes $2 + e^8 e^{2y} = e^8 e^{3y}$, or $e^{3y} - e^{2y} = 2 e^{-8} \doteq$0.6709252558e-3. Using the series expansions $e^{3y} = 1 + 3y + \cdots$ and $e^{2y} = 1 + 2y + \cdots$, we have $3y - 2y + \cdots =$0.6709252558e-3, so $y \doteq$0.6709252558e-3 and $x \doteq 4.000670925.$ This is still an approximation, but it is a pretty good one: with this new $x$ the left-hand side is 2986.960670, while the right-hand-side is 2986.964042.

10. Dec 13, 2016

### chwala

Thanks Ray, yeah right I know the solution may be approximated by various numerical methods techniques, I appreciate bro

11. Dec 14, 2016

### vela

Staff Emeritus
Those aren't negative roots; they're complex roots.

You had a cubic polynomial, so it has only three roots. The one real root gave you the solution you wanted. The two complex roots won't yield real solutions, so there are no more solutions in the (assumed) domain of the problem.

12. Dec 14, 2016

Agreed