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Math Help - One question on logs.

  1. #1
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    One question on logs.

    I need to know how to do this without a calculator.

    http://img35.imageshack.us/img35/8409/logr.jpg


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  2. #2
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     - 2\log _a (3) = \log _a \left( {3^{ - 2} } \right) = \log _a \left( {\frac{1}{9}} \right)
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  3. #3
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    Thanks, but how will I know the answers to those questions w.o my calculator?
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  4. #4
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    Quote Originally Posted by NYCKid09 View Post
    Thanks, but how will I know the answers to those questions w.o my calculator?
    You have been shown why  \log _a \left( {\frac{1}{9}} \right) = - 2\log _a (3). And the question tells you the value of \log_a (3). So what's the trouble ...?
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  5. #5
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    log base a (3) = 0.5 is the starting equation. You want to find what log base a (1/9) is.

    We can agree that one-ninth is simply 3^-2 (three to the power of negative two), for 1/3^2 would be 1/9. So now we have log base a (3^-2).

    If you don't remember this from your logarithm rules, the exponent can be placed to the left of the log. So we go from log base a (3^-2) to [-2 * log base a (3)].

    You are told that log base a (3) equals 0.5, so what's 0.5 * -2? Hopefully you know that this is -1 without needing a calculator.
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