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Math Help - Logarithms

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    Logarithms

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    MHF Contributor MarkFL's Avatar
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    Re: Logarithms

    What do you think is the best first step?
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    Re: Logarithms

    Hey Eraser147.

    Try collecting the 10^(t) terms to one side and then use the relationship that log_10(10^t) = t where log_10(x) = log(x)/log(10) where log(x) is the natural logarithm.
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    Re: Logarithms

    I ended up with 10^t+1=.8*10^t+1.6. I don't know where to go from there.
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    Re: Logarithms

    10^(t) - 0.8*10^(t) = 1.6 - 1 which means
    0.2*10^(t) = 0.6 which means
    10^(t) = 0.6/0.2 = 3.
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    Re: Logarithms

    Oh... THANKS A BUNCH. My error was not realizing I should subtract the 0.8*10^(t)
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    Re: Logarithms

    Quote Originally Posted by Eraser147 View Post
    Click image for larger version. 

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    I find this whole thread off.

    10^t+7.9=1.1(10^t)+7.7

    0.1(10^t)=0.2

    10^t=2

    t=\log_{10}(2)
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