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Math Help - question about laws of logarithms

  1. #1
    Member Jones's Avatar
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    question about laws of logarithms

    Hi,

    The laws of logarithms are pretty confusing to me.

    I have the following problem:

    \frac{1}{2^x} = \frac{5}{8^{x+3}}

    This can be rewritten as 2^{-x} = 5*8^{-(x+3)}

    if you log both sides you'll get: -xln2 = ln5*-(x+3)ln8


    so when we have a multiplication like this -(x+3)*ln8 does this imply addition?

    I don't understand the basic arithmetic operations seem to have been switched, just to maximize confusion.

    Someone please explain how this works =/

    Jones
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by Jones View Post
    Hi,

    The laws of logarithms are pretty confusing to me.

    I have the following problem:

    \frac{1}{2^x} = \frac{5}{8^{x+3}}

    This can be rewritten as 2^{-x} = 5*8^{-(x+3)}

    if you log both sides you'll get: -xln2 = ln5*-(x+3)ln8


    so when we have a multiplication like this -(x+3)*ln8 does this imply addition?

    I don't understand the basic arithmetic operations seem to have been switched, just to maximize confusion.

    Someone please explain how this works =/

    Jones
    \frac{1}{2^x} = \frac{5}{8^{x+3}}

    \Rightarrow 2^{-x} \cdot 8^{x + 3} = 5

    \Rightarrow 2^{-x} \cdot (2^3)^{x + 3} = 5

    \Rightarrow 2^{-x} \cdot 2^{3x + 9} = 5

    \Rightarrow 2^{2x + 9} = 5

    \Rightarrow \log 2^{2x + 9} = \log 5

    \Rightarrow (2x + 9) \log 2 = \log 5

    \Rightarrow 2x + 9 = \frac{\log 5}{\log 2}

    and it is simple to make x the subject. The log is to any base you want it to be, by the way.
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