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Exponential functions

  1. Oct 7, 2007 #1
    1. The problem statement, all variables and given/known data

    2^x =x^3

    show that the equation has at least two solutions.

    2. Relevant equations



    3. The attempt at a solution

    I got xln2=3lnx but I don't know what I could do from here
     
    Last edited: Oct 8, 2007
  2. jcsd
  3. Oct 8, 2007 #2
    I took a differnet approach and thought about finding values where 2^x < x^3 then where 2^x > x^3

    I found that between 1 and 3 there is a solution idk where the other one is though.
     
  4. Oct 8, 2007 #3

    morphism

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    Try x=10.
     
  5. Oct 8, 2007 #4
    i got f(2) = -4 f(0)=1 and it is continuos on this interval so there is a zero here and then

    f(4) = -16
    f(10) =24
    and it is also continous here so here is a zero

    I subtracted the RHS so the equation is 2^x - x^3 =0

    I just want to make sure this is a valid way to do this problem. And wondering if there is an easier way to pick points where I think a zero may occur other than guessing.
     
  6. Oct 8, 2007 #5
    Well.. by the problem description, all you have to show is that there is 2 or more points which the equation is true, and if you can "literally" find the points that satisfies the equation, this definitely answers the question.
     
  7. Oct 8, 2007 #6
    try this

    2^x>x^3
    you do logarithm operation on both of them

    x* log 2 >3* log x (based 10)

    x>[3/(log2)] * log x

    now i think its easier to find this number by guessing
     
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