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Thread: log and exponential questions

  1. September 8th 2008, 07:15 PM #1
    3deltat
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    log and exponential questions

    These aren't too hard... they are just some problems that Im not quite sure on. If you can help for each I really would appreciate it..

    The first one is...

    Find the limit as x goes to infinity for ( (e^3x)-(e^-3x) ) / ( (e^3x)+(e^-3x) )

    now i know you are supposed to divide num and denom by e^3x to =
    lim x -> oo ( 1-(3^-6x) ) / ( 1+(e^-6x) )
    which is basically one minus infinity over 1 plus infinity..

    The answer of course, is -1... how do i reason that? Im confused there...



    Second question is, Differentiate Log10(x/(x-1))
    The first thing i did to make myself confused is thinking that log base 10 x is = lnx.. What is the real relationship? I forgot... and the second thing, could I change the function to f(x) = log base 10 x - log base 10 (x-1) and then differentiate from there?


    And the last question is to evaluate the integral of Log base 10 x over x dx
    (sorry for having to say it in words)
    For this one, the book says something about changing it to lnx/ln10/x... i think... do you have to do it this way?


    Thank you for your help
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  2. September 8th 2008, 07:24 PM #2
    Jhevon
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    Quote Originally Posted by 3deltat View Post
    These aren't too hard... they are just some problems that Im not quite sure on. If you can help for each I really would appreciate it..

    The first one is...

    Find the limit as x goes to infinity for ( (e^3x)-(e^-3x) ) / ( (e^3x)+(e^-3x) )

    now i know you are supposed to divide num and denom by e^3x to =
    lim x -> oo ( 1-(3^-6x) ) / ( 1+(e^-6x) )
    which is basically one minus infinity over 1 plus infinity..

    The answer of course, is -1... how do i reason that? Im confused there...
    if -1 is the answer, you have your signs mixed up somewhere. this limit is 1

    Second question is, Differentiate Log10(x/(x-1))
    The first thing i did to make myself confused is thinking that log base 10 x is = lnx.. What is the real relationship? I forgot... and the second thing, could I change the function to f(x) = log base 10 x - log base 10 (x-1) and then differentiate from there?


    And the last question is to evaluate the integral of Log base 10 x over x dx
    (sorry for having to say it in words)
    For this one, the book says something about changing it to lnx/ln10/x... i think... do you have to do it this way?


    Thank you for your help
    for these two, use the change of base formula.

    recall, \log_a b = \frac {\log_c b}{\log_c a}

    in particular, \log_{10} X = \frac {\ln X}{\ln 10}


    X here is any function within the domain of the log
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  3. September 8th 2008, 07:46 PM #3
    Shyam
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    Second question is, Differentiate Log10(x/(x-1))
    The first thing i did to make myself confused is thinking that log base 10 x is = lnx.. What is the real relationship? I forgot... and the second thing, could I change the function to f(x) = log base 10 x - log base 10 (x-1) and then differentiate from there?

    {\text{We know that, }}\boxed {\log _{10} x = \frac{{\ln x}}<br /> {{\ln 10}}} \hfill \\

    f\left( x \right) = \log _{10} \left( {\frac{x}<br /> {{x - 1}}} \right) = \log _{10} x - \log _{10} \left( {x - 1} \right) \hfill \\

    f\left( x \right) = \frac{{\ln x}}<br /> {{\ln 10}} - \frac{{\ln \left( {x - 1} \right)}}<br /> {{\ln 10}} \hfill \\

    {\text{Now differentiate,}} \hfill \\

    f'\left( x \right) = \frac{1}<br /> {{x\ln 10}} - \frac{1}<br /> {{\left( {x - 1} \right)\ln 10}} \hfill \\

    = \frac{1}<br /> {{\ln 10}}\left( {\frac{1}<br /> {x} - \frac{1}<br /> {{x - 1}}} \right) \hfill \\

    = \frac{1}<br /> {{\ln 10}}\left( {\frac{{x - 1 - x}}<br /> {{x\left( {x - 1} \right)}}} \right) \hfill \\

    = \frac{{ - 1}}<br /> {{x\left( {x - 1} \right)\ln 10}} \hfill \\
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  4. September 8th 2008, 08:02 PM #4
    Shyam
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    And the last question is to evaluate the integral of Log base 10 x over x dx
    (sorry for having to say it in words)
    For this one, the book says something about changing it to lnx/ln10/x... i think... do you have to do it this way?


    Thank you for your help[/quote]

    \int {\frac{{\log _{10} x}}<br /> {x}dx} \hfill \\

    = \int {\frac{{\ln x}}<br /> {{x\ln 10}}dx} \hfill \\

    = \frac{1}<br /> {{\ln 10}}\int {\frac{{\ln x}}<br /> {x}} dx \hfill \\

    {\text{Now, let, }}\ln x = y \Rightarrow \frac{1}<br /> {x}dx = dy \hfill \\

    = \frac{1}<br /> {{\ln 10}}\int {y\;dy} \hfill \\

    = \frac{1}<br /> {{\ln 10}}\left( {\frac{{y^2 }}<br /> {2}} \right) + c \hfill \\

    = \frac{{\left( {\ln x} \right)^2 }}<br /> {{2\ln 10}} + c \hfill \\
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  5. September 8th 2008, 08:15 PM #5
    Shyam
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    [quote=3deltat;184167]These aren't too hard... they are just some problems that Im not quite sure on. If you can help for each I really would appreciate it..

    The first one is...

    Find the limit as x goes to infinity for ( (e^3x)-(e^-3x) ) / ( (e^3x)+(e^-3x) )

    now i know you are supposed to divide num and denom by e^3x to =
    lim x -> oo ( 1-(3^-6x) ) / ( 1+(e^-6x) )
    which is basically one minus infinity over 1 plus infinity..

    The answer of course, is -1... how do i reason that? Im confused there...

    The answer is 1. it is not -1.

    = \mathop {\lim }\limits_{x \to \infty } \frac{{e^{3x} - e^{ - 3x} }}<br /> {{e^{3x} + e^{ - 3x} }} \hfill \\

    {\text{Divide numerator and denominator with }}e^{3x} , \hfill \\

    = \mathop {\lim }\limits_{x \to \infty } \frac{{1 - e^{ - 6x} }}<br /> {{1 + e^{ - 6x} }} \hfill \\

    = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{1 - \frac{1}<br /> {{e^{6x} }}}}<br /> {{1 + \frac{1}<br /> {{e^{6x} }}}}} \right) \hfill \\

    = \frac{{1 - 0}}<br /> {{1 - 0}} = 1 \hfill \\
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  6. September 9th 2008, 06:55 PM #6
    3deltat
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    For the first one, if the lim was as x goes to negative infinity, would it also be one, or negative one?
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