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The Function h(x)=e^(2x)-2

  1. Oct 20, 2015 #1
    1. The problem statement, all variables and given/known data
    h(x)=e^(2x)-2
    a)Write down the Dh
    b)Solve the equation e^(2x)-2=e^x
    2. Relevant equations


    3. The attempt at a solution
    a)x∈ℝ

    e^(2x)-2=e^x
    ((e^x)^2)-e^x-2=0


    I don't know what to do here:
     
    Last edited: Oct 20, 2015
  2. jcsd
  3. Oct 20, 2015 #2

    Mark44

    Staff: Mentor

    Move the ex term to the left side. Your equation is quadratic in form, and can be factored. Keep in mind that ex > 0 for all x.
     
  4. Oct 20, 2015 #3
    Hi Mark,
    Thank you,
    I initially did this but how do I factor e^(x2)
    I also considered adding a ln to both sides to get rid of the e on both sides but the -2 is the problem.
     
  5. Oct 20, 2015 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You need to go back and review the laws of exponents. How does ##e^{2x}## relate to ##e^x##? Surely you must have had such material already, or else you would not have been asked to do this problem.
     
  6. Oct 20, 2015 #5

    Mark44

    Staff: Mentor

    What you wrote in post #1 should be something of a hint.
    If you have terms added together, taking the log is not any help.
     
  7. Oct 20, 2015 #6
    ((e^x)^2)-2=e^x
    (k^2)-k-2=0 k=e^x
    (k+1)(k-2)=0
    k=-1 or k=+2

    e^x=2
    lne^x=ln2
    x=ln2
     
  8. Oct 20, 2015 #7

    Mark44

    Staff: Mentor

    This is correct, but you should say why your are discarding the other value of k.
     
  9. Oct 21, 2015 #8
    1.a) Dh=
    b)The solution can go through this path: (e^x)^2-e^x-2=0
    If e^x=y i) then the equation is transformed like that: y^2-y-2=0
    then the solutions are y=(1±3)/2 and from i) we get e^x=(1±3)/2 BUT since the function e^x can only give positive results for x∈ℝ the only acceptable solution would be the one with the +. So e^x=2 ⇔ x=ln2, there you go friend I hope this helps :)
     
  10. Oct 21, 2015 #9

    Mark44

    Staff: Mentor

    The OP has already gotten all of this.
     
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