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Prove e^ln(a)*b = a^b

  1. Jan 31, 2016 #1
    Prove e^[ln(a)*b] = a^b

    I understand perfectly why e^ln(x) = x ... and ....I see why it works numerically but I can't justify it in terms of proof? I'd be satisfied if I could dilute this into some other proofs I'm familiar with like exponent properties such as c^(a+b) = (c^a)*(c^b) but I can't seem to figure it out intuitively.. searching the internet hasn't yielded a desired answer either.

    Edit: Nvm... e^(ab) = (e^a)^b so then (e^ln(a))^b = a^b
  2. jcsd
  3. Jan 31, 2016 #2
    It will make sense only in terms of the definition of ln(x) and exp(x) being inverse operations.

    Intuition with inverse operations and functions is a bit different from what you may be used to.

    It's more like (sin(arcsin(x)) = x
  4. Jan 31, 2016 #3
    Yea I was doing a problem that had gotten simplified down to y = (1/2) e^[(ln(10/5))*t] and couldn't figure out how they got to y= (1/2)*10^(t/5) but then I remembered x^(ab) = (x^a)^b... ln and exp make intuitive sense to me since log base e of e... I just hit a psychological blocker since I had to apply too many rules at once.
  5. Jan 31, 2016 #4


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    Start observing that ## \left(\ln{a}\right)\cdot b=\ln{a^{b}}##...
  6. Jan 31, 2016 #5
    Well the exponent has to stay inside the ln function so you end up with e^(ln(a^b)) but then I suppose a^b would drop down, so yea that would work too.
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