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Homework Help: How do I solve exponentials with base e?

  1. Sep 19, 2005 #1
    I don't know how to solve x for things like e^x=20.
    The problem I have to solve is x^3+e^(2x)+8=0
    Can anyone help, please?
  2. jcsd
  3. Sep 19, 2005 #2
    have you learned logarithms?

    what is the natural log of e^x?
  4. Sep 19, 2005 #3


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    You won't be able to find an exact solution to your problem so you will have to resort to solving it graphically or by iteration.
  5. Sep 19, 2005 #4


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    The exponential function is the inverse function of the natural logarithm.
    therefore [tex] ln(e^{f(x)}) = f(x)[/tex]. In your example, like any equation, what you do on one side of the equation must be done on the other side. You can take the natural log of both sides [tex]ln(e^{x})=ln(20) [/tex]
    Then using the property I gave above you should be able to solve for x

    Once you have practised using the idea given above, you can then tackle your second question. You may want to bring your x-terms on one side of equation and any other terms, to the other side of equation. As with the first query, take the natural log of both sides. Though you eliminate your exponential function, you may still have some natural log terms left. That's okay, by substitution of values for x you should arrive at correct value f(x) to your question.

    Edit: By plotting the values of x you substitute and resulting values f(x), as Tide suggests; by observing the trend in your graph, you may find (or at least narrow down) your solutions more quickly.
    Last edited: Sep 19, 2005
  6. Sep 22, 2005 #5
    You mean I just replace values for x until I get that the equation equals 0. That's what I did first and got aproximately -2.001. I was just wondering if I could solve the equation instead of replacing numbers.
  7. Sep 22, 2005 #6
    Anytime you have a logarithm, there are different rules that you need to know. For the first example you gave the one you need to know is this.

    [tex]\log_{a}b^c=c\log_{a}b[/tex] it also follows that

    [tex]\ln a^b=b\ln a[/tex]

    So when you are given something like [tex]e^x=20[/tex], how can you apply this rule to solve that?

    For your second problem, as others have said, you cannot solve for x explicity so another method must be used.
  8. Sep 22, 2005 #7
    it is equal to x,

    well, I just wanted to remind theOP
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