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Simple exponential function exercise

  1. Sep 6, 2011 #1
    1. The problem statement, all variables and given/known data

    is e^x^2 = 4 equivalent to e^x = 2

    2. Relevant equations

    As above

    3. The attempt at a solution

    This is just an exercise, but I'm quite stuck as how to show this is true (or false for that matter). I thought to take the log of both sides and use the log identity to get rid of the double exponent and cancel out the 'ln e' (=1):

    ln e^x^2 = ln 4

    x^2 ln e = ln 4

    x^2 = ln 4

    but I'm not sure this helps me (kinda went around in a circle!), nor am I clear on another method to use. I'm sure there's a log property I'm missing? :S
  2. jcsd
  3. Sep 6, 2011 #2


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    e^(x^2) and (e^x)^2 are two different things. e^x^2 doesn't mean much on its own. You probably mean (e^x)^2=4. Try taking ln of that.
  4. Sep 6, 2011 #3
    Oh my mistake, I should have been clear, it's e^(x^2) = 4 equiv to e^x = 2.
  5. Sep 7, 2011 #4


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    I think you are asking does e(x2) = (ex)2

    Let x=some appropriate number, say, 3, and use your calculator.

    Re-examining, ex2 = (ex)2
    you are really asking does x2 = 2x

    Well, it does if x=2 or 0 :frown:
  6. Sep 7, 2011 #5
    I have to learn how to TEX >.<

    The question is to show an equivalence between equation 1 and equation 2 where:

    equation 1: e^(x^2) = 4 (so that's e to the x-squared equals 4)

    equation 2: e^(x) = 2 (and this is e to the x equals 2)

    So I think I have to manipulate equation 1 into the form of equation 2. To be honest, I haven't ever seen a question like this, we didn't do it in class, it's on an (unmarked) exercise quiz relating to basic algebra skills (which I blatantly lack, I keep gettng tripped up on stuff like this, just when I think I understand something!)

    I should mention it's a TRUE or FALSE question, so they may not be equivalent, but I'd like to know! :P
  7. Sep 7, 2011 #6


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    If the value of x that solves eqn 2 also solves eqn 1 (using your calculator), then there's a good bet that it is TRUE. Try that for starters.

    Your restatement of the eqns amounts to what I wrote, in any case.
  8. Sep 7, 2011 #7


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    Last edited by a moderator: Apr 26, 2017
  9. Sep 7, 2011 #8
    OK so I did this:

    e^x = 2
    ln e^x = ln 2 (log law ln e^x = x ln e)
    x ln e = ln 2 (ln e = 1, cancels out)
    x = ln 2

    and sub that into the first equation and sure enough I get 4, so they are equivalent. Now I feel like a right duffer! Thanks for the pointers though, I'm just getting a feel for logs and I get a bit flustered when I don't quite grasp the question.

  10. Sep 7, 2011 #9


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    I disagree. (e^x)^2= e^(2x) which is an easier way to write it. I would immediately assume that e^(x^2) was meant.

    And if (e^x)^2= 4 was meant, I would NOT take the logarithm. Since the "outer" function is squaring, I would take the square root first: e^x= +/- 2. Since an exponential (of a real number) cannot be negative, e^x= 2, x= ln(2) is the only (real) solution.
  11. Sep 7, 2011 #10

    Ray Vickson

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    No, they are not. The solutions of exp(x^2)=4 are x = +-sqrt(2*ln(2)) = +- 1.1774, while the solution of exp(x)=2 is x = ln(2) = 0.6931. It IS true that x^2 = 2x for some special values of x, but not for those values that solve either of your two equations.

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