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Math Help - Help with logs

  1. #1
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    Help with logs

    Could somebody please help a middle aged student who hasn't done any 'proper' maths for 25 years (and is starting to remember why he preferred English lessons) with this problem where I need to express with x as the subject? I have the answer but mine doesn't match it

    Many thanks

    z=e^{x^2}{^/}{^2}
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  2. #2
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    Well, let's learn one more thing...

    Please show your work.

    You obviously know a logarithm is needed, so I'll give you this step.

    ln(z)\;=\;log_{e}(z)\;=\;\frac{x^{2}}{2}

    Now what?
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  3. #3
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    OK so

    x=\sqrt2log_{e}z

    Which is the correct answer. How though did you get to that first stage? What am I not seeing - apart from the blindingly obvious, before you say it.
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  4. #4
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    Quote Originally Posted by cistudent View Post
    OK so

    x=\sqrt2log_{e}z

    Which is the correct answer. How though did you get to that first stage? What am I not seeing - apart from the blindingly obvious, before you say it.
    If a^x = b then  x = \log_a{b}

    This formula is used to convert index form to log form.
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  5. #5
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    I must be stupid. I'm still not getting it. Can you explain as if you were telling this to somebody who started his 'log chapter' today. Which I did, I hadn't even heard of e before this afternoon.

    Thanks
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  6. #6
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    Quote Originally Posted by cistudent View Post
    I must be stupid. I'm still not getting it. Can you explain as if you were telling this to somebody who started his 'log chapter' today. Which I did, I hadn't even heard of e before this afternoon.

    Thanks
    OK. I hope you know {a^m}\time{a^n} = a^(m+n)
    and \frac{a^m}{a^n} = a^(m-n)

    10^0 = 1

    10^1 = 10

    Then 5 = 10^x, where value of x will be some where between 0 and 1.

    In the calculator find log(5), you will see 0.6990.

    That means 10^(0.6990) = 5.

    It is written as log_{10}{5} = 0.6990
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  7. #7
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    Nope! I'm still banging my head against the wall here. Could somebody take me through it step by step and explain each step. It's after 3.00am here and I can't sleep until I get this clear in my head
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  8. #8
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    Take the log of both sides. log(e)^a = a

    log and e^ and inverses of each other.

    log(z) = log(e^(x^2/2))= (x^2)/2
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  9. #9
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    By definition, a logarithm is the inverse of an exponential. So logarithms undo exponentials and exponentials undo logarithms.

    In other words:

    If y = a^x then \log_a{y} = x where a is any base.


    Proof: y = a^x

    \log_a{y} = \log_a{a^x}

    \log_a{y} = x (since logarithms undo exponentials).


    Similarly:

    If y = \log_a{x} then a^y = x where a is any base.


    Proof: y = \log_a{x}

    a^y = a^{\log_a{x}}

    a^y = x (since exponentials undo logarithms).



    Now, looking at your question.

    z = e^{\frac{x^2}{2}}.

    In this case our base is the number e, Euler's Number. It pops up so often that you should research it...

    We want to undo the exponential, so we use a logarithm of base e.


    z = e^{\frac{x^2}{2}}

    \log_e{z} = \log_e{e^{\frac{x^2}{2}}}

    \log_e{z} = \frac{x^2}{2} since the logarithm undoes the exponential

    2\log_e{z} = x^2

    x = \pm\sqrt{2\log_e{z}}.


    Note: Since the logarithm of base e occurs so often in nature, it is called the Natural Logarithm, and is often denoted as \ln.

    So it may be that the answer is written as

    x = \pm\sqrt{2\ln{z}}.
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  10. #10
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    Ah now I see. So log_{e}e=1

    It's so bloomin obvious, why didn't I see that.

    This was the last of a list of exercises and that whole e^x{^2}{^/}{^2} thing threw me and made it seem more complicated than it really was, when all I had to do was bring down the whole exponent and multiply by one.

    Thanks very much Niall and prove it. Now I can finally get to bed.
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