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Integrating a complex ln

  1. Apr 16, 2012 #1
    1. The problem statement, all variables and given/known data

    integrate:13((4^x)+(3^x))dx

    2. Relevant equations



    3. The attempt at a solution

    I know the solution is 13((4^x)/ln(4) + (3^x)/ln(3)) + C

    Can someone explain to me how this works? I don't know where the ln's are coming from. How would I differentiate this back to the original function?
     
  2. jcsd
  3. Apr 17, 2012 #2

    cepheid

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    Are you aware of the exponential function ex which can be defined by the the following property?[tex]\frac{d}{dx}e^x = e^x[/tex]Given that, consider something like 4x. By the definition of the natural logarithm and its inverse, the exponential function, you can write "4" as eln4. Therefore, our function becomes:[tex]4^x = (e^{\ln4})^x = e^{x\ln4} [/tex]It follows from the chain rule that[tex]\frac{d}{dx}(4^x) = \frac{d}{dx}(e^{x\ln4}) = \ln4 e^{x\ln4} = (\ln4)4^x [/tex]Now, in this example it didn't matter that the number was 4. It could have been anything. So, using the properties of the exponential function and the natural logarithm, we have shown that, for any number "a", it is true that:[tex]\frac{d}{dx}a^x = (\ln a) a^x [/tex]
     
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