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Thread: Logarithmic Help

  1. July 17th 2008, 04:50 PM #1
    Maga22
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    Logarithmic Help

    I tried the following problem:

    log (log x) = 1

    I need to know if I did it right.

    Problem 1:

    log(log x) = 1

    log x = e^1
    log x = e

    ANSWER: x = e^e

    The second problem:

    3^log x = 3x

    I attempted but got stuck. Any help will be much appreciated.

    3^log x = 3x

    log(3^log x) = log 3x
    log3logx = log3+logx
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  2. July 17th 2008, 05:06 PM #2
    Jhevon
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    Quote Originally Posted by Maga22 View Post
    I tried the following problem:

    log (log x) = 1

    I need to know if I did it right.

    Problem 1:

    log(log x) = 1

    log x = e^1
    log x = e

    ANSWER: x = e^e
    yes

    The second problem:

    3^log x = 3x

    I attempted but got stuck. Any help will be much appreciated.

    3^log x = 3x

    log(3^log x) = log 3x
    log3logx = log3+logx
    nice start. now just treat log(x) as a variable and solve for it. what if you have (log3)y = log3 + y. could you solve for y? sure you could. here, of course, y = logx

    we get y = \frac {\log 3}{\log 3 - 1}

    so, \log x = \frac {\log 3}{\log 3 - 1}

    can you finish?
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  3. July 18th 2008, 01:08 AM #3
    Maga22
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    Logarithmic Help

    Thanks for the feed back on problem one. I am still stuck on the second problem. This is as far as I got:

    Problem: 3^logx=3x

    log(3^logx)=log3x
    log 3 log x=log 3+log x
    (logx)^2=log 3+log x
    (logx)^2-log x-log 3=0
    (logx)^2-logx ??

    Can you help please? I'm a little confused with the sample provided. I attempted again, but not sure if I am on the right track. I have several more problems like this and I need to have some sort of sample to go off of in order to start and complete the others. Thanks.
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  4. July 18th 2008, 01:16 AM #4
    flyingsquirrel
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    Hi
    Quote Originally Posted by Maga22 View Post
    Thanks for the feed back on problem one. I am still stuck on the second problem. This is as far as I got:

    Problem: 3^logx=3x

    log(3^logx)=log3x
    log 3 log x=log 3+log x
    (logx)^2=log 3+log x
    Where does (\log x)^{\color{red}2} come from ?

    From \log 3 \log x =\log 3+\log x we get, subtracting \log x to both sides, \log 3 \log x -\log x=\log 3

    Factor : (\log 3 -1)\log x =\log 3

    Can you take it from here ?
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  5. July 18th 2008, 01:17 AM #5
    Chris L T521
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    Quote Originally Posted by Maga22 View Post
    I tried the following problem:

    log (log x) = 1

    I need to know if I did it right.

    Problem 1:

    log(log x) = 1

    log x = e^1
    log x = e

    ANSWER: x = e^e
    Quote Originally Posted by Jhevon View Post
    yes
    Sorry man, but I disagree. e^{\log(\log(x))}\neq\log(x) However, 10^{\log(\log(x))}=\log(x) since \log(x)=\log_{\color{red}10}(x). If the problem had contained \ln(x) or \log_{\color{red}e}(x), then the answer would be x=e^e...

    Thus,

    10^{\log(\log(x))}=10^{1}

    \implies\log(x)=10

    \implies \color{red}\boxed{x=10^{10}}

    That should be the correct answer.


    Quote Originally Posted by Maga22 View Post
    The second problem:

    3^log x = 3x

    I attempted but got stuck. Any help will be much appreciated.

    3^log x = 3x

    log(3^log x) = log 3x
    log3logx = log3+logx
    Quote Originally Posted by Jhevon View Post
    nice start. now just treat log(x) as a variable and solve for it. what if you have (log3)y = log3 + y. could you solve for y? sure you could. here, of course, y = logx

    we get y = \frac {\log 3}{\log 3 - 1}

    so, \log x = \frac {\log 3}{\log 3 - 1}

    can you finish?
    Good. I have no problems with this.

    --Chris
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  6. July 18th 2008, 01:29 AM #6
    Moo
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    Hello Chris,

    I think it depends on what you intend by \log. I've seen several websites in English where \log is used for base e. Though most of them use it for base 10...
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  7. July 18th 2008, 01:29 AM #7
    Maga22
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    Quote Originally Posted by flyingsquirrel View Post
    Hi

    Where does (\log x)^{\color{red}2} come from ?

    From \log 3 \log x =\log 3+\log x we get, subtracting \log x to both sides, \log 3 \log x -\log x=\log 3

    Factor : (\log 3 -1)\log x =\log 3

    Can you take it from here ?

    I divided both sides by log 3-1 and got this



    This where I got stuck earlier. Can someone please assist?
    Last edited by Maga22; July 18th 2008 at 01:39 AM. Reason: missing info
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  8. July 18th 2008, 01:33 AM #8
    Chris L T521
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    Quote Originally Posted by Moo View Post
    Hello Chris,

    I think it depends on what you intend by \log. I've seen several websites in English where \log is used for base e. Though most of them use it for base 10...
    Moo, you have a point. In America, I was taught that \log(x)=\log_{10} (x), and that \ln(x)=\log_e(x). However, as you mentioned, there are other ways it can be interpreted [i.e. \log(x)=\log_e(x)]. I'll keep that in mind as other problems like this pop up.

    --Chris
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  9. July 18th 2008, 01:49 AM #9
    flyingsquirrel
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    Quote Originally Posted by Maga22 View Post
    I divided both sides by log 3-1 and got this



    This where I got stuck earlier. Can someone please assist?
    If \log is the natural logarithm (i.e. \exp(\log x )=x) then \log x=\frac{\log 3}{\log 3-1}\implies \underbrace{\exp (\log x)}_{x}=\exp\left( \frac{\log 3}{\log 3-1}\right)\implies x=\exp\left( \frac{\log 3}{\log 3-1}\right)

    If \log is the base 10 logarithm then 10^{\log x}=x and, exactly as in the previous case, we get x=10^{\frac{\log 3}{\log 3-1}}
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