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Homework Help: Possibly impossible question?

  1. Jan 1, 2010 #1
    Given that 2e^(t²) + e^t - 3e = 0, solve for t.

    I've been racking my brains to figure out a solution to this one. I know that this can easily be solved by using a graphing calculator, but is there an analytic method to deduce the value of t as well? According to my friend, since this is neither a polynomial nor an algebraic expression, an analytic method would be impossible (at least at A levels).

    Thanks in advance to all those who help!
    Last edited: Jan 1, 2010
  2. jcsd
  3. Jan 1, 2010 #2


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    Try taking the natural logarithm of both sides then using the quadratic equation on the resulting second degree polynomial
  4. Jan 1, 2010 #3
    But that would only work if the first term is to the power 2t instead of t².
  5. Jan 1, 2010 #4
    Is your equation:
    [tex]2(e^t)^2 +e^t - 3e = 0[/tex]
    [tex]2e^{(t^2)} + e^t -3e = 0[/tex]
    ? I suspect the former as that's a very common type of problem in elementary algebra and in that case you can indeed get by using the substitution y=e^t.
  6. Jan 1, 2010 #5
    The latter. If it were the former I wouldn't even be posting it here :p
  7. Jan 1, 2010 #6
    obviously, one solution is t = 1.
  8. Jan 1, 2010 #7


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    The natural logarithm is 0 is not defined, and the logarithm of a sum is not the sum of the logarithms, so this does not work at all.
  9. Jan 2, 2010 #8


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