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Math Help - another logarithm problem

  1. #1
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    Question another logarithm problem

    can someone help me with this one?

    4^3p*4^2p-1=100 the answer is .8644 thanks!!

    i need someone to show me to break it down step by step.

    i did 4^3p*4^2p-1 =4^5p-1 is that right or would it be=16^5p-1?
    Last edited by Evandmal07; May 8th 2009 at 08:41 PM.
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  2. #2
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    Hello Evandmal07
    Quote Originally Posted by Evandmal07 View Post
    can someone help me with this one?

    4^3p*4^2p-1=100 the answer is .8644 thanks!!

    i need someone to show me to break it down step by step.

    i did 4^3p*4^2p-1 =4^5p-1 is that right or would it be=16^5p-1?
    You were right first time: 4^{3p} \times 4^{2p-1} = 4^{5p-1}.

    The key to solving this type of problem is the log formula:

    • \log(a^b)=b\log (a)

    So if you now take logs of both sides (base 10 is easiest):

    \log(4^{5p-1})=\log(100) = \log(10^2) = 2\log(10) =2

    \Rightarrow (5p-1)\log 4= 2

    \Rightarrow5p-1 =\frac{2}{\log 4}

    \Rightarrow p =\frac15\Big(\frac{2}{\log 4}+1\Big)=0.8644 (to 4 d.p.)

    Grandad
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  3. #3
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    thank you!!!

    thank you so much now i see!!! i just didnt realize that i needed to convert that log100 to 2 but i see that that makes sense!!

    wow thanks so much!!!
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