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Thread: 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; Aug 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,
    $\displaystyle \log_b(xy)=\log_b x+\log_b y$
    Thus,
    $\displaystyle \log_9 (x-8)+\log_9 x$
    Becomes,
    $\displaystyle \log_9 [x(x-8)]=1$
    By the defintion of logarithms,
    $\displaystyle 9^1=x(x-8)$
    Thus,
    $\displaystyle x^2-8x=9$
    Thus,
    $\displaystyle x^2-8x-9=0$
    Thus,
    $\displaystyle (x-9)(x+1)=0$
    Thus,
    $\displaystyle x=9,-1$
    But,
    $\displaystyle x\not =-1$ because logarithms are not definied for negative numbers. Thus, $\displaystyle x=9$

    4^4 logbase4^4...

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


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

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

    Thanks a lot, again.
    I think you mean,
    $\displaystyle (6/7)^x=343/216$, right?
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  3. #3
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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.
    $\displaystyle log_9(x-8) + log_9(x) = 1$

    $\displaystyle log_9(x^2-8x) = 1$

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

    Thus x = 9, -1. However note that x = -1 gives an impossibility: $\displaystyle log_9(-1-8) = log_9(-9)$ 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:
    $\displaystyle (6/7)^{x+1} = 343/216$

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

    Thus
    $\displaystyle x + 1 = -3$

    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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