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Math Help - help needed logs problem

  1. #1
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    help needed logs problem

    log5 (2x+3) = log5 (3) log to base 5
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  2. #2
    Super Member bigwave's Avatar
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    If the bases are the same

    the definition is

    log_a{a}^b = b

    so log_5{5}^1 = 1

    log_{5}5 (2x+3) = log_5{5}(3)

    then (2x+3) = 3

    x = 0
    Last edited by bigwave; December 21st 2009 at 07:30 PM.
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  3. #3
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    right now im just practicing for my exam

    logx + log (x-21) = 2
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  4. #4
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    Hello math321
    Quote Originally Posted by math321 View Post
    right now im just practicing for my exam

    logx + log (x-21) = 2
    I assume that these are logs to base 10; in which case:
    \log x +\log(x-21) = 2

    \Rightarrow \log x(x-21)= \log 10^2

    \Rightarrow x^2-21x=100

    \Rightarrow (x-25)(x+4)=0

    \Rightarrow x =25, -4
    Grandad
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  5. #5
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    \log x +\log(x-21) = 2

    \Rightarrow \log x(x-21)= \log 10^2

    \Rightarrow x^2-21x=100

    \Rightarrow (x-25)(x+4)=0

    \Rightarrow x =25, \textcolor{red}{-4}

    ... be sure to check solutions in the original equation to avoid domain issues.

    ...
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