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Math Help - Logarithmic equations

  1. #1
    dto
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    Exclamation Logarithmic equations

    Question: Solve teh logarithmic equation. Be sure to reject any value that is not in the domain of the original logarithmic expressions. Give the exact answers

    (Log 2^(x+2)) + (log 2^(x-4))=4

    I know the answer is 6, but is completely clueless on how to get 6.

    here is what my work looks like:

    = log 2^(x+2)(x-4)=4

    = log 2 ^ (x^2-2x-8)=4

    = log 2^(x^2-2x-12)

    = quadratic formula

    = x=2 +/- square root of 52 / 2

    that is where i stopped, because 52 is not a perfect square.

    Please help ASAP!

    Test is tomorrow!
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by dto View Post
    Question: Solve teh logarithmic equation. Be sure to reject any value that is not in the domain of the original logarithmic expressions. Give the exact answers

    (Log 2^(x+2)) + (log 2^(x-4))=4

    I know the answer is 6, but is completely clueless on how to get 6.

    here is what my work looks like:

    = log 2^(x+2)(x-4)=4

    = log 2 ^ (x^2-2x-8)=4

    = log 2^(x^2-2x-12)

    = quadratic formula

    = x=2 +/- square root of 52 / 2

    that is where i stopped, because 52 is not a perfect square.

    Please help ASAP!

    Test is tomorrow!
    I believe you made a mistake with the laws of exponents.

    Also I don't understand your notation. Do you mean \log_2 (x+2) or \log_{10}\, (2^{x+2})



    ====================

    Edit I'm going to assume the former because f(6) = 4 for it.

    \log_2 (x+2) + \log_2 (x-4) = 4

    \log_2 [(x+2)(x-4)] = \log_2 (x^2-2x-8) = 4

    x^2-2x-8 = 2^4 = 16

    x^2-2x-24=0

    Solve the quadratic - this one will factorise
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