1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solving an exponential function

  1. Dec 12, 2016 #1
    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. jcsd
  3. Dec 12, 2016 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes, you can do it.
     
  4. Dec 13, 2016 #3

    i will post solution soon, i dont have calculator now but i know solution will be of the form ## x=ln p##
     
  5. Dec 13, 2016 #4
    the solution is thus ##p=54.64→x=ln 54.64, ⇒x=4.00##
     
  6. Dec 13, 2016 #5

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes, very close to it.
     
  7. Dec 13, 2016 #6
    yes, ##x=4.00 ## is an approximate solution to the problem.
     
  8. Dec 13, 2016 #7
    Might there be other approximate solutions to the problem, i ignored the negative values of ##p=-0.0183+1.41i, -0.0183-1.41i##
     
  9. Dec 13, 2016 #8

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  10. Dec 13, 2016 #9

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  11. Dec 13, 2016 #10
    Thanks Ray, yeah right I know the solution may be approximated by various numerical methods techniques, I appreciate bro
     
  12. Dec 14, 2016 #11

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
  13. Dec 14, 2016 #12
    Agreed
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Solving an exponential function
  1. Exponential function (Replies: 4)

  2. Exponential function (Replies: 1)

Loading...