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Thread: Simple but messed up

  1. #1
    Member roshanhero's Avatar
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    Simple but messed up

    In my book the integral of a^x is directly given as $\displaystyle \int a^xdx=\frac{a^x}{log a}$
    Can anyone tell me how can we derive it?
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  2. #2
    Member Black's Avatar
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    Use the identity $\displaystyle a^x=e^{\text{log}(a)x}$.
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  3. #3
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    Quote Originally Posted by roshanhero View Post
    In my book the integral of a^x is directly given as $\displaystyle \int a^xdx=\frac{a^x}{log a}$
    Can anyone tell me how can we derive it?
    let $\displaystyle y = a^x$, then $\displaystyle \ln y = x \ln a$ and $\displaystyle \frac{1}{y} \frac{dy}{dx} = \ln a$, or $\displaystyle dx = \frac{1}{y} \frac{dy}{\ln a} $

    and the integral became

    $\displaystyle \int a^xdx=\int y \frac{1}{y} \frac{dy}{\ln a} = \frac{a^x}{\ln a}$
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  4. #4
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    Hi , First of All take your book away. it is not a mathematics , maybe it is

    Literature book .

    Solve The Integral By This Way :

    $\displaystyle

    \int a^x dx = \int \left({e^{\ln a}}\right)^x dx = \int {{e^{(\ln a)x}}} dx
    $

    $\displaystyle
    u=x\ln a \rightarrow du = dx\ln a
    $

    $\displaystyle
    \int e^u \frac {du}{\ln a}=\ln a\int e^u = \frac {1}{\ln a} {a^x}
    $

    Note :

    $\displaystyle

    e^u = e^{x \ln a} =\left( e^{\ln a} \right)^{x} = a^x

    $
    Last edited by mr fantastic; Jan 15th 2010 at 02:29 AM. Reason: Fixed latex
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  5. #5
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    Quote Originally Posted by roshanhero View Post
    In my book the integral of a^x is directly given as $\displaystyle \int a^xdx=\frac{a^x}{log a}$
    Can anyone tell me how can we derive it?
    You're expected to learn something from the help you get in other threads. In this instance I refer specifically to http://www.mathhelpforum.com/math-he...questions.html
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