Solving an exponential function

[chwala]

Homework Statement

[/B]
Solve the equation
$e^{2x}+2=e^{3x-4}$

Homework Equations

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

 

[ehild]

Homework Statement

[/B]
Solve the equation
$e^{2x}+2=e^{3x-4}$

Homework Equations

The Attempt at a Solution

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

Yes, you can do it.

 

[chwala]

Yes, you can do it.

i will post solution soon, i don't have calculator now but i know solution will be of the form $x=ln p$

 

[chwala]

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

 

[ehild]

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

Yes, very close to it.

 

[chwala]

Yes, very close to it.

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

 

[chwala]

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

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

 

[Ray Vickson]

Homework Statement

[/B]
Solve the equation
$e^{2x}+2=e^{3x-4}$

Homework Equations

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

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.

 

[Ray Vickson]

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

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.

 

[chwala]

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.

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

 

[vela]

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

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.

 

[chwala]

Agreed