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Math Help - Logarithms

  1. #1
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    Logarithms

    Final exam's tomorrow, and I have a lot of trouble with this type of a problem:



    I would think that the setup involves something with logs, but that's about all I know. Thanks in advance!
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  2. #2
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    Quote Originally Posted by deltoids
    Final exam's tomorrow, and I have a lot of trouble with this type of a problem:



    I would think that the setup involves something with logs, but that's about all I know. Thanks in advance!
    You have,
    4^{x+3}=5^{x+2}
    Take the log of both sides,
    \log 4^{x+3}=\log 5^{x+2}
    Thus, by exponent rules,
    (x+3)\log 4=(x+2)\log 5
    Thus,
    x\log 4+3\log 4=x\log 5+2\log 5
    Thus,
    x\log 4-x\log 5=2\log 5-3\log 4
    Thus,
    x(\log 4-\log 5)=2\log 5-3\log 4
    Thus,
    x=\frac{2\log 5-3\log 4}{\log 4-\log 5}\approx 4.21
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  3. #3
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    Much Appreciated sir!

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