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Math Help - 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: 1 = \log (x + 1) - \log x

    now recall: \log X - \log Y = \log \frac XY and \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: 1 = \log (x + 1) - \log x

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

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

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

    10x = x+1

    9x = 1

    x = \frac 19

    Oh, awesome. Thanks, I simply haven't thought about transforming the log into a 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...

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

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

    10x = x+1

    9x = 1

    x = \frac 19

    Oh, awesome. Thanks, I simply haven't thought about transforming the log into a 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

    <br />
\log x + 1 = \log (x + 1)

    into

    <br />
\log x + \log 10 = \log (x + 1)

    assuming we are in base ten.

    then using the rule for addition of logs we get,

    <br />
\log 10x = \log (x + 1)

    removing the log from both sides we get

    <br />
10x = x + 1 <br />

    and thus getting


     x = \frac{1}{9}


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