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Math Help - Logarithms

  1. #1
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    Question Logarithms

    I am stuck on explaining how to use the definition to find the derivative of lnyx in the following situation. I have to define the area under a curve
    y=1/x in the terms of the natural logarithm function. wThe area under the curve f(x) is defined as the integral f(x) dx, if f(x)=1/x. Then the area under the curve is log1/x dx which equals ln lxl. So where do i go from here as far as explaining how to use the definition to find the derivative of lnyx? Any help?

    On a related note, how do would you use the derivative formula from above to prove ln(a)+ln(b)= ln(ab)?

    Thank you for any help in advance.
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by bemidjibasser View Post
    I am stuck on explaining how to use the definition to find the derivative of lnyx in the following situation. I have to define the area under a curve
    y=1/x in the terms of the natural logarithm function. wThe area under the curve f(x) is defined as the integral f(x) dx, if f(x)=1/x. Then the area under the curve is log1/x dx which equals ln lxl. So where do i go from here as far as explaining how to use the definition to find the derivative of lnyx? Any help?

    On a related note, how do would you use the derivative formula from above to prove ln(a)+ln(b)= ln(ab)?

    Thank you for any help in advance.
    Consider the INTEGRAL (1/t) dt on [1, x]; so f(t) is 1/t. The fundamental theorem of calculus then says:
    If F is the antiderivative of f, then the definite integral is equal to
    F(x) - F(1) = ln x - ln 1 = ln x

    So ln x = INTEGRAL (1/t) dt on [1, x].

    The ln yx is a bit confusing. Could that be a typo?
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  3. #3
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    apcalculus- Thanks for the response. I typed what was in the question that I am working from. Not sure either...any suggestions on how to use the definition of the derivative in general to find the derivative of ln of ??? Again, any help is appreciated.
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  4. #4
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    Quote Originally Posted by bemidjibasser View Post
    apcalculus- Thanks for the response. I typed what was in the question that I am working from. Not sure either...any suggestions on how to use the definition of the derivative in general to find the derivative of ln of ??? Again, any help is appreciated.
    Say, y= ln x. Take the exp(y)=x --- 1
    Take the derivative of exp(y) wrt x.

    You would have exp(y)*y' = 1
    therefore, y'=1/exp(y)=1/x (from 1)

    I don't know if this is what you are looking for though.
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  5. #5
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    Apluscalculus- It was a typo. It should read y= ln x. Sorry about that.
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  6. #6
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    So, I think I have this part right...

    y= ln x, so x=e^y which gives 1=e^y dy/dx , then 1/e^y =dy/dx, which gives1/e^lnx =dy/dx , and lastly dy/dx=1/x
    Right? Now, how would I be able to use this to prove ln(a) + ln(b) = ln(ab)?
    Thanks again for the help.
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  7. #7
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    Quote Originally Posted by bemidjibasser View Post
    So, I think I have this part right...

    y= ln x, so x=e^y which gives 1=e^y dy/dx , then 1/e^y =dy/dx, which gives1/e^lnx =dy/dx , and lastly dy/dx=1/x
    Right? Now, how would I be able to use this to prove ln(a) + ln(b) = ln(ab)?
    Thanks again for the help.
    Yep, you got it right.
    2) You might just use the inverse of ln i.e. by taking exp on both sides and applying power rule.
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