Results 1 to 9 of 9

Math Help - logarithm equation!! need help solving...

  1. #1
    Newbie
    Joined
    Dec 2010
    Posts
    18

    logarithm equation!! need help solving...

    The questions asks me to find the relationship between x and y by removing the 'log10' in the equation

    2log10y + 3log10x = 2

    The answer is supposedly y = 10x^3/2

    I keep getting up to log10y = 3log10x but don't know how to continue on from there...

    Please help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    \displaystyle log_{10}y^2+log_{10}x^3=2\Rightarrow log_{10}(y^2*x^3)=2\Rightarrow 10^{log_{10}(y^2*x^3)}=10^2\Rightarrow\cdots
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2010
    Posts
    18
    sorry, i get your answer right up to the last step.....how do i get from '10^log10(y^2*x^3) = 10^2' to y = ....??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    \displaystyle 10^{log_{10}x}=x

    It is the inverse.

    \displaystyle e^{ln(x)}=x Same concept. This is in base e.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2010
    Posts
    18
    um...so =>

    y^2*x^3 = 10^2
    y^2 = 10^2/x^3

    so y=√(〖10〗^2/x^3 )

    ?????

    im really sorry...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by dwsmith View Post
    \displaystyle log_{10}y^2+log_{10}x^3=2\Rightarrow log_{10}(y^2*x^3)=2\Rightarrow 10^{log_{10}(y^2*x^3)}=10^2\Rightarrow\cdots
    \displaystyle y^2*x^3=10^2\Rightarrow y^2=\frac{10^2}{x^3}\Rightarrow y=\frac{10}{x^{3/2}}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Dec 2010
    Posts
    18
    okay, i get it now (thanks!!)
    just one problem...the textbook says the answer is 10*x^3/2 not 10/x^3/2
    is that the books mistake or am i missing something?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    It should say then \displaystyle y=10x^{-3/2}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Dec 2010
    Posts
    18
    probably....thanks again for your time!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. solving logarithm
    Posted in the Algebra Forum
    Replies: 2
    Last Post: December 2nd 2010, 04:00 AM
  2. Help with solving this logarithm
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 2nd 2010, 09:08 PM
  3. Solving for x in a logarithm
    Posted in the Algebra Forum
    Replies: 3
    Last Post: March 21st 2009, 06:53 AM
  4. Need help solving this logarithm
    Posted in the Algebra Forum
    Replies: 4
    Last Post: December 17th 2008, 05:34 AM
  5. Solving a logarithm equation
    Posted in the Algebra Forum
    Replies: 9
    Last Post: May 22nd 2006, 11:51 AM

Search Tags


/mathhelpforum @mathhelpforum