Results 1 to 8 of 8

Math Help - Logs

  1. #1
    Junior Member
    Joined
    Oct 2007
    Posts
    56

    Unhappy Logs

    How to you answer the following question using logs:

    What are the two solutions for x of 2^(2x)-8*2^x+15=0
    In words: 2 to the power 2x, -8 multiplied by 2 to the power x, plus 15.

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

  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by haku View Post
    How to you answer the following question using logs:

    What are the two solutions for x of 2^(2x)-8*2^x+15=0
    In words: 2 to the power 2x, -8 multiplied by 2 to the power x, plus 15.

    Please help!
    2^{2x} - 8*2^x + 15 =(2^x)^2 - 8*2^x + 15=0
    let \, u=2^x

    then this is just a quadratic equation in u, that is

    u^2 - 8u +15 = (u-3)(u-5)=0

    so u=3 or u=5
    but since

     u= 2^x

    therefore x=\frac{\ln 3}{\ln 2} = \log_2 3 \, or \, x=\frac{\ln 5}{\ln 2} = \log_2 5
    Last edited by kalagota; October 31st 2007 at 07:06 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2007
    Posts
    56

    Thumbs up

    Thanks for the help and your explanation
    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
    From 2^x=3\implies x=\log_23 & 2^x=5\implies x=\log_25.

    From this way, it's easier to check the solutions without a calculator.

    --

    2^{2x}-8\cdot2^x+15=0.

    So for x=\log_23,

    2^{2x}=2^{\log_29}=9 & 2^x=3, therefore

    9-8\cdot3+15=0.

    You can do the same procedure to verify the second solution.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2007
    Posts
    56
    What about questions in the form:
    log(base6)4 + log(base6)9 + log(base3)45 -log(base3)5

    Thanks
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by haku View Post
    What about questions in the form:
    log(base6)4 + log(base6)9 + log(base3)45 -log(base3)5

    Thanks
    It would be log(base 6) 36 + log(base 3) 9
    Can you take it from here? Just to help you, the answer should be 4.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Oct 2007
    Posts
    56
    I understand that part but what do you do when the base in a different value?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by haku View Post
    I understand that part but what do you do when the base in a different value?
    what i usually do is, to let a variable to the sum of logs with same base and another variable for the other base, then i add them after solving independently..
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: February 22nd 2011, 06:39 PM
  2. Logs
    Posted in the Algebra Forum
    Replies: 4
    Last Post: April 24th 2010, 08:52 AM
  3. Logs
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 10th 2009, 07:08 PM
  4. Dealing with Logs and Natural Logs
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 14th 2008, 07:18 AM
  5. several questions-logs/natural logs
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 12th 2007, 09:58 PM

Search Tags


/mathhelpforum @mathhelpforum