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Math Help - Natural log

  1. #1
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    Natural log

    Define the natural logarithm of a number x in terms of an integral.Show that
    ln(xy) = ln(x) + ln(y)
    for all x,y >0
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by matty888 View Post
    Define the natural logarithm of a number x in terms of an integral.Show that
    ln(xy) = ln(x) + ln(y)
    for all x,y >0
    \ln(x)\equiv\int_1^{x}\frac{1}{y}dy
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by matty888 View Post
    Define the natural logarithm of a number x in terms of an integral.Show that
    ln(xy) = ln(x) + ln(y)
    for all x,y >0
    define \ln(x)=\int_{1}^{x}\frac{1}{t}dt

    we want to show \ln(xy)=\ln(x)+\ln(y)

    so we start with

    \ln(xy)=\int_{1}^{xy}\frac{1}{t}dt

    using the linear property of integrals we get

    \int_{1}^{xy}\frac{1}{t}dt=\int_{1}^{x}\frac{1}{t}  dt+\int_{x}^{xy}\frac{1}{t}dt

    lets focus on the 2nd integral

    \int_{x}^{xy}\frac{1}{t}dt

    let u=\frac{t}{x} \iff ux =t

    du=\frac{1}{x}dt \iff xdu=dt

    \int_{x}^{xy}\frac{1}{t}dt=\int_{1}^{y}\frac{1}{ux  }xdu=\int_{1}^{y}\frac{1}{u}du=\ln(y)

    so finally with all the above work

    \ln(xy)=\int_{1}^{xy}\frac{1}{t}dt=\int_{1}^{x}\fr  ac{1}{t}dt+\int_{1}^{y}\frac{1}{u}du=\ln(x)+\ln(y)

    Yeah!!
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  4. #4
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    thank you!!

    Thank you very much TheEmptySet,you made it very clear and understandable,if you could help me with some of my other post i would be very grateful
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by matty888 View Post
    Define the natural logarithm of a number x in terms of an integral.Show that
    ln(xy) = ln(x) + ln(y)
    for all x,y >0
    Isnt this legal for the proof?

    \ln(xy)=\ln\bigg(e^{\ln(x)}e^{\ln(y)}\bigg)=\ln\bi  gg(e^{\ln(x)+\ln(y)}\bigg)=ln(x)+\ln(y)

    This just popped into my head on the way home? Its so easy...but is it mathematically legal?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Isnt this legal for the proof?

    \ln(xy)=\ln\bigg(e^{\ln(x)}e^{\ln(y)}\bigg)=\ln\bi  gg(e^{\ln(x)+\ln(y)}\bigg)=ln(x)+\ln(y)

    This just popped into my head on the way home? Its so easy...but is it mathematically legal?
    yes it is
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Jhevon View Post
    yes it is
    Haha...sweet..I thought it was easy
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  8. #8
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    Quote Originally Posted by Mathstud28 View Post
    Its so easy...but is it mathematically legal?
    No it is not. How is e^x defined? In fact, e^x is defined to be the inverse function of \ln x but you never demonstrated it.
    ---
    Here is how to prove it. Let y>0. Define a function 0,\infty)\mapsto \mathbb{R}" alt="f0,\infty)\mapsto \mathbb{R}" /> as f(x) = \ln (xy) - \ln x. This function is differenciable and furthermore, f'(x) = \frac{y}{yx} - \frac{1}{x} = 0. Thus, f(x) = k for some number k. In particular, f(1) = k but f(1) = \ln (y) - \ln (1) = \ln y. Thus, k=\ln y. Therefore, \ln (xy) - \ln x = \ln y.
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by Mathstud28 View Post
    Isnt this legal for the proof?

    \ln(xy)=\ln\bigg(e^{\ln(x)}e^{\ln(y)}\bigg)=\ln\bi  gg(e^{\ln(x)+\ln(y)}\bigg)=ln(x)+\ln(y)

    This just popped into my head on the way home? Its so easy...but is it mathematically legal?
    No because you have to use the definition in terms of the integral, you do not know any other definition for the purposes of this question. That is you need a proof like that of TheEmptySet's post.

    RonL
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