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Thread: Logarithm problem

  1. #1
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    Logarithm problem

    I'm having difficulties solving the following equation:

    (logx)+1 = log(x+1)

    Any help would be greatly appreciated. Thanks.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Sodapop View Post
    I'm having difficulties solving the following equation:

    (logx)+1 = log(x+1)

    Any help would be greatly appreciated. Thanks.
    note that you have: $\displaystyle 1 = \log (x + 1) - \log x$

    now recall: $\displaystyle \log X - \log Y = \log \frac XY$ and $\displaystyle \log_a b = c \implies a^c = b$
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    note that you have: $\displaystyle 1 = \log (x + 1) - \log x$

    now recall: $\displaystyle \log X - \log Y = \log \frac XY$ and $\displaystyle \log_a b = c \implies a^c = b$
    Hmm, then...

    $\displaystyle \log \frac {x+1}{x} = 1$

    $\displaystyle 10^1 = \frac {x+1}{x}$

    $\displaystyle 10x = x+1$

    $\displaystyle 9x = 1$

    $\displaystyle x = \frac 19$

    Oh, awesome. Thanks, I simply haven't thought about transforming the log into a $\displaystyle 10^x$ form. You're the best.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Sodapop View Post
    Hmm, then...

    $\displaystyle \log \frac {x+1}{x} = 1$

    $\displaystyle 10^1 = \frac {x+1}{x}$

    $\displaystyle 10x = x+1$

    $\displaystyle 9x = 1$

    $\displaystyle x = \frac 19$

    Oh, awesome. Thanks, I simply haven't thought about transforming the log into a $\displaystyle 10^x$ form. You're the best.
    yup, that's it. good job
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  5. #5
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    Hey there Sodapop

    I would've made

    $\displaystyle
    \log x + 1 = \log (x + 1) $

    into

    $\displaystyle
    \log x + \log 10 = \log (x + 1) $

    assuming we are in base ten.

    then using the rule for addition of logs we get,

    $\displaystyle
    \log 10x = \log (x + 1) $

    removing the log from both sides we get

    $\displaystyle
    10x = x + 1
    $

    and thus getting


    $\displaystyle x = \frac{1}{9}$


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