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Exponential integration

  1. Feb 9, 2013 #1
    1. The problem statement, all variables and given/known data

    exponential integration formula [itex]∫a^{u} du = \frac {a^{u}}{ln(a)} +c [/itex]

    ∫ [itex]\left(\frac{2}{3}\right)^{x} dx[/itex]

    2. Relevant equations


    3. The attempt at a solution

    ∫ [itex]\left(\frac{2}{3}\right)^{x} dx[/itex]

    = [itex]\frac{\left(2/3 \right)^{x}}{ln(2)/ln(3)}[/itex] <<< this is my answer ???

    = - [itex]\frac{\left(2/3 \right)^{x}}{ln(3)/ln(2)}[/itex] << this answer from wolframalpha why ln(3)/ln(2) not ln(2)/ln(3) and where the minus comes from

    please advise

    thank you
    Last edited: Feb 9, 2013
  2. jcsd
  3. Feb 9, 2013 #2
    It's just logarithm rules:


    Btw, you can compare if two expressions are equivalent in Wolfram with " -(2/3)^x/(ln(3/2))==(2/3)^x/(ln(2/3)) ". As you can see, the output is "True".

    EDIT: I'm assuming you meant ln(3/2) and ln(2/3) instead of ln(3)/ln(2) and ln(2)/ln(3), as the former is 1) correct and 2) what Wolfram actually gives as an answer.
    Last edited: Feb 9, 2013
  4. Feb 9, 2013 #3
    thank you
  5. Feb 9, 2013 #4
    None of the answers you have mentioned is correct. You have done the same mistake again as you did in your previous threads.

    [tex]\ln \frac{a}{b}≠\frac{\ln a}{\ln b}[/tex]
  6. Feb 9, 2013 #5


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    Both of the following answers are incorrect.
    Actually, WolframAlpha gives:

    [itex]\displaystyle \int \left(\frac{2}{3}\right)^{x} dx=-\frac{\left(2/3 \right)^{x}}{\ln(3/2)}+\text{constant}\ .[/itex]

    As for the minus sign: [itex]\displaystyle \ \ \ln\left(\frac{2}{3}\right)=\ln\left(\frac{3}{2} \right)^{-1}\!\!=(-1)\ln\left(\frac{3}{2}\right)\ .[/itex]

    I see Pranav-Arora beat me to it!   (Way to go, P-A !)

    Added in Edit:

    By the way:

    [itex]\displaystyle \ln\left(\frac{2}{3}\right)=\ln(2)-\ln(3) [/itex]

    On the other hand: [itex]\displaystyle \ \ \frac{\ln(2)}{\ln(3)}=\log_{\,3}(2) \ .[/itex]
    Last edited: Feb 9, 2013
  7. Feb 9, 2013 #6

    ohhh I have to be more careful about this

    Thank you Pranav-Arora

    Thank you Sammy for the clarification
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