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Thread: Logarithmic Help

  1. #1
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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. #2
    is up to his old tricks again! Jhevon's Avatar
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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 $\displaystyle y = \frac {\log 3}{\log 3 - 1}$

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

    can you finish?
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  3. #3
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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. #4
    Super Member flyingsquirrel's Avatar
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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 $\displaystyle (\log x)^{\color{red}2}$ come from ?

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

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

    Can you take it from here ?
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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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. $\displaystyle e^{\log(\log(x))}\neq\log(x)$ However, $\displaystyle 10^{\log(\log(x))}=\log(x)$ since $\displaystyle \log(x)=\log_{\color{red}10}(x)$. If the problem had contained $\displaystyle \ln(x)$ or $\displaystyle \log_{\color{red}e}(x)$, then the answer would be $\displaystyle x=e^e$...

    Thus,

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

    $\displaystyle \implies\log(x)=10$

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

    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 $\displaystyle y = \frac {\log 3}{\log 3 - 1}$

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

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

    --Chris
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  6. #6
    Moo
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    Hello Chris,

    I think it depends on what you intend by $\displaystyle \log$. I've seen several websites in English where $\displaystyle \log$ is used for base e. Though most of them use it for base 10...
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  7. #7
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    Quote Originally Posted by flyingsquirrel View Post
    Hi

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

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

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

    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; Jul 18th 2008 at 01:39 AM. Reason: missing info
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Moo View Post
    Hello Chris,

    I think it depends on what you intend by $\displaystyle \log$. I've seen several websites in English where $\displaystyle \log$ 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 $\displaystyle \log(x)=\log_{10} (x)$, and that $\displaystyle \ln(x)=\log_e(x)$. However, as you mentioned, there are other ways it can be interpreted [i.e. $\displaystyle \log(x)=\log_e(x)$]. I'll keep that in mind as other problems like this pop up.

    --Chris
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  9. #9
    Super Member flyingsquirrel's Avatar
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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 $\displaystyle \log$ is the natural logarithm (i.e. $\displaystyle \exp(\log x )=x$) then $\displaystyle \log x=\frac{\log 3}{\log 3-1}\implies \underbrace{\exp (\log x)}_{x}=\exp\left( \frac{\log 3}{\log 3-1}\right)\implies x=\exp\left( \frac{\log 3}{\log 3-1}\right)$

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