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Math Help - Logs...

  1. #1
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    Logs...

    You guys were pretty good last time, so I wanted to see what your thoughts were on these logs...

    I have to find x for this equation...

    logbase9*(x-8) + logbase9 x = 1

    do I have 2x? And then, how do I get rid of logbase9?

    This one also ****** me off, 4^4 logbase4^4...

    so it's 4, to the 4logbase4, to another power of 4... errr...

    last is...

    (6/7)^x+1 = 343/216

    How do I get rid of that (6/7) ?

    Thanks a lot, again.
    Last edited by ThePerfectHacker; August 3rd 2006 at 04:35 PM.
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  2. #2
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    Quote Originally Posted by juangohan9
    You guys were pretty good last time, so I wanted to see what your thoughts were on these logs..

    I have to find x for this equation...

    logbase9*(x-8) + logbase9 x = 1
    Use the rule,
    \log_b(xy)=\log_b x+\log_b y
    Thus,
    \log_9 (x-8)+\log_9 x
    Becomes,
    \log_9 [x(x-8)]=1
    By the defintion of logarithms,
    9^1=x(x-8)
    Thus,
    x^2-8x=9
    Thus,
    x^2-8x-9=0
    Thus,
    (x-9)(x+1)=0
    Thus,
    x=9,-1
    But,
    x\not =-1 because logarithms are not definied for negative numbers. Thus, x=9

    4^4 logbase4^4...

    so it's 4, to the 4logbase4, to another power of 4... errr...
    Easy 4^4=256
    So you have,
    256\log_4 4^4
    Now what is? \log 4^4
    Use the rule,
    \log_b b^n=n
    Thus,
    256 \cdot 4= 1024


    (6/7)^x+1 = 343/216

    How do I get rid of that (6/7) ?

    Thanks a lot, again.
    I think you mean,
    (6/7)^x=343/216, right?
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by juangohan9
    You guys were pretty good last time, so I wanted to see what your thoughts were on these logs...

    I have to find x for this equation...

    logbase9*(x-8) + logbase9 x = 1

    do I have 2x? And then, how do I get rid of logbase9?

    This one also ****** me off, 4^4 logbase4^4...

    so it's 4, to the 4logbase4, to another power of 4... errr...

    last is...

    (6/7)^x+1 = 343/216

    How do I get rid of that (6/7) ?

    Thanks a lot, again.
    log_9(x-8) + log_9(x) = 1

    log_9(x^2-8x) = 1

    Thus
    x^2 - 8x = 9^1 = 9
    x^2 - 8x - 9 = (x-9)(x+1) = 0

    Thus x = 9, -1. However note that x = -1 gives an impossibility: log_9(-1-8) = log_9(-9) in the original equation is undefined. Thus x = 9 is the only solution.

    I'm not quite sure what you mean by "4^4 logbase4^4." I can't make your description of it make sense. Is it a typo or am I seeing something wrong here?

    I presume the last one is supposed to be:
    (6/7)^{x+1} = 343/216

    Note that 6^3 = 216 and 7^3 = 343. Thus
    (6/7)^{x+1} = (7/6)^3 = (6/7)^{-3}

    Thus
    x + 1 = -3

    and x = -4.

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Hmmm...TPH and I seem to be busy with all the same posts today...

    -Dan
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