Results 1 to 5 of 5

Thread: Math 12 logarithms

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    16

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    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) $\displaystyle \log_a(a^{2x})=log_b(b^{3x-3})$.

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

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

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

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

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

    We can't do any more as the bases of the logarithms are not the same.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    Hold on there, Pardner.

    $\displaystyle log_{1/x}(a) = \frac{log(a)}{log(1/x)} = \frac{log(a)}{-log(x)} = -log_{x}(a)$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    This looks about right:

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2008
    Posts
    16
    Hey, thanks guys!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: May 29th 2011, 12:57 AM
  2. Replies: 2
    Last Post: Feb 14th 2011, 02:31 AM
  3. Replies: 1
    Last Post: Jun 19th 2010, 10:05 PM
  4. Replies: 1
    Last Post: Jun 2nd 2010, 09:49 AM
  5. Replies: 2
    Last Post: Feb 15th 2010, 03:09 PM

Search Tags


/mathhelpforum @mathhelpforum