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Math Help - Math 12 logarithms

  1. #1
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    Math 12 logarithms

    Hi all,

    I just a few questions on Math 12, not sure how to do these questions (I need to know the how-to more than the answer itself):

    1) "The population of a type of bacteria triples every 20 hours. In how many hours will a population of 30 become a population of 1000?"

    2) "Solve for x: log(subcript a) a ^ 2x = log (subscript b^2) b^(3x-3)


    3) Simplify: [log (subscript: 1/x) 1/y ] - [log (subscript 1/x) y ] - [log (subscript X) 1/y]

    (This is from the Logarithms unit).


    Thanks a lot in advance!
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  2. #2
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    Quote Originally Posted by funnytim View Post
    Hi all,

    I just a few questions on Math 12, not sure how to do these questions (I need to know the how-to more than the answer itself):

    1) "The population of a type of bacteria triples every 20 hours. In how many hours will a population of 30 become a population of 1000?"

    2) "Solve for x: log(subcript a) a ^ 2x = log (subscript b^2) b^(3x-3)


    3) Simplify: [log (subscript: 1/x) 1/y ] - [log (subscript 1/x) y ] - [log (subscript X) 1/y]

    (This is from the Logarithms unit).


    Thanks a lot in advance!
    2) \log_a(a^{2x})=log_b(b^{3x-3}).

    Do you remember the rule \log_a(a^m)=m?

    So we get 2x=3x-3. Can you solve for x?

    3) \log_{\frac{1}{x}}(\frac{1}{y})-\log_{\frac{1}{x}}(y)-\log_x(\frac{1}{y})

    Do you remember the rule \log_a(m)-\log_a(n) = \log_a(\frac{m}{n})? Using this we get

    \log_{\frac{1}{x}}(\frac{\frac{1}{y}}{y})-\log_x(\frac{1}{y})
    =\log_{\frac{1}{x}}(\frac{1}{y^2})-\log_x(\frac{1}{y})

    We can't do any more as the bases of the logarithms are not the same.
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  3. #3
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    Hold on there, Pardner.

    log_{1/x}(a) = \frac{log(a)}{log(1/x)} = \frac{log(a)}{-log(x)} = -log_{x}(a)
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  4. #4
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    This looks about right:

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  5. #5
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    Hey, thanks guys!
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