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Math Help - log help - test tomarrow

  1. #1
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    log help - test tomarrow

    im not sure how to type logs so try to bare with me >< first set of parenthesis is the base of the log
    log(4)x + log(4)(x-2)= log(4)8
    if you could explain that i would be super happy
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  2. #2
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    Hello, Porcelain!

    \log_4(x) + \log_4(x-2) \:=\:\log_4(8)
    On the left side, combine the logs . . .

    We have: . \log_4\bigg[x(x-2)\bigg] \:=\:\log_4(8)

    "Un-log" both sides: . x(x-2) \:=\:8 \quad\Rightarrow\quad x^2 - 2x - 8 \:=\:0

    . . Then: . (x-4)(x+2)\:=\:0 \quad\Rightarrow\quad x \:=\:4,-2

    But x = -2 is an extraneous root.


    . . Therefore, the solution is: . \boxed{x \:=\:4}

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  3. #3
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    Quote Originally Posted by Porcelain View Post
    im not sure how to type logs so try to bare with me>< first set of parenthesis is the base of the log
    log(4)x + log(4)(x-2)= log(4)8
    if you could explain that i would be super happy
    \log_4{x} + \log_4(x-2) = \log_4{8}

    \log_4[x(x-2)] = \log_4{8}

    x(x-2) = 8

    solve the quadratic ... don't forget to check both solutions in the original equation for domain issues.
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  4. #4
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    so if a log of the same base is equal to a log of the same base you get rid of the logs?
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  5. #5
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    Log a + Log b = Log (a*b) and yes you get rid of the logs if they have same base
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