Results 1 to 2 of 2

Thread: Precalculus

  1. May 21st 2009, 11:54 AM #1
    juicysharpie
    juicysharpie is offline
    Junior Member
    Joined
    Jan 2009
    Posts
    74

    Question Precalculus

    If
    <br /> 3^{a} = 4^{b} = 36

    Find
    (2/a) + (1/b).

    I'm studying for a test, and I don't understand this problem at all. Thanks for the help!
    Follow Math Help Forum on Facebook and Google+
  2. May 21st 2009, 12:22 PM #2
    TheEmptySet
    TheEmptySet is offline
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by juicysharpie View Post
    If
    <br /> 3^{a} = 4^{b} = 36

    Find
    (2/a) + (1/b).

    I'm studying for a test, and I don't understand this problem at all. Thanks for the help!

    3^{a}=36 \iff a=\frac{\ln(36)}{\ln(3)}

    4^{b}=36 \iff b=\frac{\ln(36)}{\ln(4)}

    \frac{2}{\frac{\ln(36)}{\ln(3)}}+\frac{1}{\frac{\l n(36)}{\ln(4)}}=\frac{2\ln(3)}{\ln(36)}+\frac{\ln( 4)}{\ln(36)}

    =\frac{2\ln(3)+\ln(4)}{\ln(36)}=\frac{\ln(9)+\ln(4 )}{\ln(36)}=\frac{\ln(36)}{\ln(36)}=1
    Follow Math Help Forum on Facebook and Google+
« Slopes of Secants | Increasing Function »

Similar Math Help Forum Discussions

  1. Precalculus
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 22nd 2009, 01:20 PM
  2. Precalculus
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 22nd 2009, 01:50 AM
  3. Precalculus
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 22nd 2009, 01:01 AM
  4. Precalculus
    Posted in the Pre-Calculus Forum
    Replies: 0
    Last Post: May 26th 2008, 10:01 AM
  5. Need PreCalculus Help
    Posted in the Algebra Forum
    Replies: 4
    Last Post: February 1st 2006, 07:26 AM

Search Tags


/mathhelpforum @mathhelpforum