Results 1 to 4 of 4

Math Help - logarithm help

  1. #1
    Newbie
    Joined
    Aug 2008
    Posts
    2

    logarithm help

    I need help with this problem. I have trouble using log to isolate x.

    (4)(3^x) = (7)(5^x)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Given: (4)(3^x) = (7)(5^x)

    Take the log of both sides: \log{[(4)(3^x)]} = \log{[(7)(5^x)]}

    Property of multiplication in logarithms: \log{4} + \log{3^x} = \log{7} + \log{5^x}<br />

    Subtract \log{4} from both sides: \log{3^x} = \log{7} + \log{5^x} - \log{4} <br />

    Subtract \log{5^x} from both sides: \log{3^x} -  \log{5^x} = \log{7} - \log{4}

    Combine the two logs on the LHS: \log{\left(\frac{3}{5}\right)^x} = \log{7} - \log{4}

    Carry down x (property of logarithm): x \log{\left(\frac{3}{5}\right)} = \log{7} - \log{4}

    Divide by \log{\left(\frac{3}{5}\right)} on both sides: x = \frac{\log{7} - \log{4}}{\log{\left(\frac{3}{5}\right)}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by student4200 View Post
    I need help with this problem. I have trouble using log to isolate x.

    (4)(3^x) = (7)(5^x)
    do you know the rules of logs?

    \log 4(3^x) = \log 7(5^x) .................log both sides

    \Rightarrow \log 4 + \log 3^x = \log 7 + \log 5^x ...............since \log_a XY = \log_a X + \log_a Y

    \Rightarrow \log 4 + x \log 3 = \log 7 + x \log 5 ............since \log_a X^n = n \log_a X

    now what?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    4\cdot 3^{x}=7\cdot 5^{x}\implies \left( \frac{3}{5} \right)^{x}=\frac{7}{4}\,\therefore\, x=\frac{\ln 7-\ln 4}{\ln 3-\ln 5}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: July 18th 2011, 08:04 AM
  2. logarithm
    Posted in the Algebra Forum
    Replies: 3
    Last Post: July 17th 2010, 05:23 AM
  3. Logarithm #2
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: June 1st 2010, 06:29 AM
  4. Logarithm
    Posted in the Algebra Forum
    Replies: 6
    Last Post: September 17th 2009, 11:23 PM
  5. Logarithm
    Posted in the Algebra Forum
    Replies: 4
    Last Post: March 13th 2008, 11:05 AM

Search Tags


/mathhelpforum @mathhelpforum