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Math Help - Relationship between a^x and log to the base a, integration.

  1. #1
    Boz
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    Relationship between a^x and log to the base a, integration.

    Could you please help me with the following problem:

    Describe the relationship between a^x and log to the base a, giving the exponential function as an example. Simplify the function
    \frac{e^{sin2x} lne^{3x}}{{(e^{cosx})}^{2sinx}e^{-x}}

    and hence evaluate the integral

    \int^1_0 \frac{e^{sin2x} lne^{3x}}{{(e^{cosx})}^{2sinx}e^{-x}}dx

    Thank you in advance
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  2. #2
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    Quote Originally Posted by Boz View Post
    Could you please help me with the following problem:

    Describe the relationship between a^x and log to the base a, giving the exponential function as an example. Simplify the function
    \frac{e^{sin2x} lne^{3x}}{{(e^{cosx})}^{2sinx}e^{-x}}

    and hence evaluate the integral

    \int^1_0 \frac{e^{sin2x} lne^{3x}}{{(e^{cosx})}^{2sinx}e^{-x}}dx

    Thank you in advance
    note that ...

    \sin(2x) = 2\sin{x}\cos{x}

    \ln(e^{3x}) = 3x i.e. ... \log_a(a^x) = x

    you will end up with the definite integral \displaystyle \int_0^1 3x \cdot e^x \, dx which can be evaluated using integration by parts.
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  3. #3
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    Quote Originally Posted by Boz View Post
    Could you please help me with the following problem:

    Describe the relationship between \displaystyle   a^x and log to the base a, giving the exponential function as an example. Simplify the function
    \displaystyle \frac{e^{sin2x} lne^{3x}}{{(e^{cosx})}^{2sinx}e^{-x}}

    and hence evaluate the integral

    \displaystyle \int^1_0 \frac{e^{sin2x} lne^{3x}}{{(e^{cosx})}^{2sinx}e^{-x}}dx
    Is that not easier to read?
    Please learn to use \displaystyle
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    Quote Originally Posted by Boz View Post
    Could you please help me with the following problem:

    Describe the relationship between a^x and log to the base a, giving the exponential function as an example.
    log_a(x) and a^x are inverse functions: log_a(a^x)= x and a^{log_a(x)}= x

    Simplify the function
    \frac{e^{sin2x} lne^{3x}}{{(e^{cosx})}^{2sinx}e^{-x}}
    ln(e^3x)= 3x, (e^{cos x})^{2sin x}= e^{2sin x cos x}= e^{sin 2x} so \frac{e^{sin 2x}lne^{3x}}{(e^{cosx})^{2sinx}e^{-x}}= \frac{e^{sin 2x}3x}{e^{sin 2x}}e^{-x}= 3xe^{-x}

    Integrate using integration by parts: u= e^{-x}, dv= 3x.

    and hence evaluate the integral

    \int^1_0 \frac{e^{sin2x} lne^{3x}}{{(e^{cosx})}^{2sinx}e^{-x}}dx

    Thank you in advance [/QUOTE]
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  5. #5
    Boz
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    Quote Originally Posted by hallsofivy View Post
    log_a(x) and a^x are inverse functions: log_a(a^x)= x and a^{log_a(x)}= x


    ln(e^3x)= 3x, (e^{cos x})^{2sin x}= e^{2sin x cos x}= e^{sin 2x} so \frac{e^{sin 2x}lne^{3x}}{(e^{cosx})^{2sinx}e^{-x}}= \frac{e^{sin 2x}3x}{e^{sin 2x}}e^{-x}= 3xe^{-x}

    integrate using integration by parts: u= e^{-x}, dv= 3x.

    And hence evaluate the integral

    \int^1_0 \frac{e^{sin2x} lne^{3x}}{{(e^{cosx})}^{2sinx}e^{-x}}dx

    thank you in advance
    [/quote]

    Here is what I did. Is it correct?

     \displaystyle \int 3xe^x  dx

    Let  u=3x, \frac{du}{dx}=3,  du=3dx,<br />
Let ,dv=e^x dx, v=\int e^x dx=e^x
     \displaystyle 3xe^x-\int e^x 3dx,<br />
3xe^x-3e^x=3e^x(x-1)
    Hence  \displaystyle  \int_0^1 3xe^x dx= [3e^x(x-1)]<br />
0 to 1= [3e^1(1-1)]-[3e^0(0-1)]=-3*(-1)=3
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