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Math Help - Please Help Me Solve These Log Equations!

  1. #1
    jwhite09
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    Please Help Me Solve These Log Equations!

    1. log 3 + log (5x - 2) = log (9x + 6)

    2. 2 log (2x - 1) = 2 log (3x + 1) - log 4

    3. log_2 (x - 4) + log_2 (x - 3) = 1

    Please Help!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by jwhite09 View Post
    1. log 3 + log (5x - 2) = log (9x + 6)
    Look at the left hand side:

    log 3 + log (5x - 2) = log(3(5x-2)) = log(15x-6),


    so:

    15x-6 = 9x+6.

    Now try the same sort of thing on the remainder.

    RonL
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  3. #3
    Bar0n janvdl's Avatar
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    Quote Originally Posted by jwhite09 View Post

    1. log 3 + log (5x - 2) = log (9x + 6)
    Log (3 \times (5x - 2)) = Log (9x + 6)

     15x - 6 = 9x + 6

    6x = 12

    x = 2


    Quote Originally Posted by jwhite09 View Post

    2. 2 log (2x - 1) = 2 log (3x + 1) - log 4
    log (2x-1)^2 = log (3x + 1)^2 - log (4)

    log (2x-1)^2 = log \frac{(3x + 1)^2}{4}

    (2x - 1)^2 = \frac{(3x + 1)^2}{4}

    4(2x - 1)^2 = (3x + 1)^2

    4(4x^2 - 4x + 1) = 9x^2 + 6x + 1

    16x^2 - 16x + 4 = 9x^2 + 6x + 1

    7x^2 - 22x + 3 = 0

    You can solve for x here, I'm lazy.


    Quote Originally Posted by jwhite09 View Post

    3. log_2 (x - 4) + log_2 (x - 3) = 1
    log_2 ((x - 4)(x - 3)) = 1

    log_2 (x^2 - 7x + 12) = 1

    2^1 = x^2 - 7x + 12

    0 = x^2 - 7x + 10

    0 = (x - 5)(x - 2)

    x = 5 \ or \ x = 2
    Last edited by janvdl; October 2nd 2007 at 11:10 AM. Reason: Typo
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