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Thread: logarithm

  1. #1
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    logarithm

    Question:Show that $\displaystyle log_{a}b=\frac{1}{log_{b}a} $,
    a) using the result $\displaystyle log_{a}b \times log_{b}c = log_{a}c $ ]b) from first principles
    I was able to do part a;

    $\displaystyle log_{a}b=\frac{log_{a}C}{log_{b}C}=\frac{log_{c}b} {log_{c}a}= \frac{log_{b}b}{log_{b}a}= \frac{1}{log_{b}a}$

    but part b is what i am struggling with. For me first principles means
    $\displaystyle \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} $
    and i have tried and gone no way please help
    Last edited by topsquark; Mar 26th 2019 at 06:51 AM. Reason: Tweaked LaTeX
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  2. #2
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    Re: logarithm

    "From first principals" just means "from the definition".

    In the context of derivatives, you use the definition you stated.

    This question though has nothing to do with derivatives.

    Can you state the derivative of a logarithm?
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  3. #3
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    Re: logarithm

    Quote Originally Posted by Debsta View Post
    "From first principals" just means "from the definition".

    In the context of derivatives, you use the definition you stated.

    This question though has nothing to do with derivatives.

    Can you state the derivative of a logarithm?
    I cannot state the derivative of a logarithm as I have not learnt it yet.
    You states earlier that from first principles means from definition.
    thank you I know what it means
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  4. #4
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    Re: logarithm

    Quote Originally Posted by bigmansouf View Post
    I cannot state the derivative of a logarithm as I have not learnt it yet.
    You states earlier that from first principles means from definition.
    thank you I know what it means
    Oh I'm sorry, my mistake. What I meant was:

    Can you state the definition of a logarithm?
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  5. #5
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    Re: logarithm

    If $\displaystyle x= log_a(b)$ then $\displaystyle b= a^x$. From that, $\displaystyle b^{1/x}= a$. Now take the logarithm, base b, of both sides of that.
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  6. #6
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    Re: logarithm

    Quote Originally Posted by bigmansouf View Post
    Question:Show that $\displaystyle log_{a}b=\frac{1}{log_{b}a} $,
    a) using the result $\displaystyle log_{a}b \times log_{b}c = log_{a}c $ ]b) from first principles
    I was able to do part a;
    [tex] log_{a}b=\frac{log_{a}C}{log_{b}C}=\frac{log_{c}b} {log_{c}a}= \frac{log_{b}b}{log_{b}a}= \frac{1}{log_{b}a
    but part b is what i am struggling with. For me first principles means $\displaystyle \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} $
    and i have tried and gone no way please help
    I admit that I am at all sure what your are asking. I assume you have been asked to derive the derivative of $\log(x)$ using $\displaystyle \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$
    I also assume that L'Hopital's Rule can be used.
    $ \begin{align*}\displaystyle \lim_{h\rightarrow 0}\frac{\log(x+h)-\log(x)}{h}&=\displaystyle \lim_{h\rightarrow 0}\frac{\log\left[\frac{x+h}{x}\right]}{h} \\&=\displaystyle \lim_{h\rightarrow 0}\frac{\log\left[1+\frac{h}{x}\right]}{h}\\\text{L'Hopital }D_h&= \lim_{h\rightarrow 0}\frac{\frac{1}{x}}{\left[1+\frac{h}{x}\right]}\\&=\frac{1}{x} \end{align*}$
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    Re: logarithm

    Quote Originally Posted by Debsta View Post
    Oh I'm sorry, my mistake. What I meant was:

    Can you state the definition of a logarithm?
    the definition for me is basically the foundation which is

    $\displaystyle a^x=b \Leftrightarrow log_{a}b=x $ from here, there is the properties of logarithm

    I did this
    part a)
    $\displaystyle log_{a}b \times log_{b}c = log_{a}c $
    $\displaystyle log_{a}b= \frac{log_{a}c}{log_{b}c}=x $
    $\displaystyle a^x=b $
    take logs to base b
    $\displaystyle log_{b}a^x = log_{b}b $
    $\displaystyle xlog_{b}a = log_{b}b $
    $\displaystyle x = \frac{log_{b}b}{log_{b}a}= \frac{1}{log_{b}a} $
    $\displaystyle x = \frac{1}{log_{b}a} $

    part b)
    from first principles
    $\displaystyle a^{x}=b $
    take logs to base b
    $\displaystyle log_{b}a^x = log_{b}b $
    $\displaystyle xlog_{b}a = log_{b}b $
    $\displaystyle x = \frac{log_{b}b}{log_{b}a}= \frac{1}{log_{b}a} $
    $\displaystyle x = \frac{1}{log_{b}a} $
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  8. #8
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    Re: logarithm

    Quote Originally Posted by Plato View Post
    I admit that I am at all sure what your are asking. I assume you have been asked to derive the derivative of $\log(x)$ using $\displaystyle \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$
    I also assume that L'Hopital's Rule can be used.
    $ \begin{align*}\displaystyle \lim_{h\rightarrow 0}\frac{\log(x+h)-\log(x)}{h}&=\displaystyle \lim_{h\rightarrow 0}\frac{\log\left[\frac{x+h}{x}\right]}{h} \\&=\displaystyle \lim_{h\rightarrow 0}\frac{\log\left[1+\frac{h}{x}\right]}{h}\\\text{L'Hopital }D_h&= \lim_{h\rightarrow 0}\frac{\frac{1}{x}}{\left[1+\frac{h}{x}\right]}\\&=\frac{1}{x} \end{align*}$
    the problem is that i dont even know what the differentiation method you use is and it is not even in the book i am studying from. As you wrote in post #5 that is what the question asked for part b

    i wanted to to know if in maths you are ask from first principles it has to depend on the context. In a sense, that if the phrase 'first principles' is used in a calculus question then it is about calculus and if not in this case logarithm then it is about logarithm

    thanks for your help
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