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Thread: 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 $\displaystyle a^x$ 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

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

    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 $\displaystyle a^x$ 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

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

    Thank you in advance
    note that ...

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

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

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

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

    and hence evaluate the integral

    $\displaystyle \displaystyle \int^1_0 \frac{e^{sin2x} lne^{3x}}{{(e^{cosx})}^{2sinx}e^{-x}}dx$
    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 $\displaystyle a^x$ and log to the base a, giving the exponential function as an example.
    $\displaystyle log_a(x)$ and $\displaystyle a^x$ are inverse functions: $\displaystyle log_a(a^x)= x$ and $\displaystyle a^{log_a(x)}= x$

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

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

    and hence evaluate the integral

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

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


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

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

    And hence evaluate the integral

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

    thank you in advance
    [/quote]

    Here is what I did. Is it correct?

    $\displaystyle \displaystyle \int 3xe^x dx$

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