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Math Help - I can't figure out this log problem!!!!

  1. #1
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    I can't figure out this log problem!!!!




    what would the x be? i am so confused...
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  2. #2
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     \log_2x+ \log_2(x+3) = \log_28

    addition of logs gives

     \log_2(x(x+3)) = \log_28

     \log_2(x^2+3x) = \log_28

    raising both sides to the power of 2

     x^2+3x = 8

     x^2+3x - 8=0

    now we have a quadratic, do you know how to solve this?
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  3. #3
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    yes ofcourse! 1.7 i believe
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  4. #4
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    Quote Originally Posted by badshahkhan View Post
    yes ofcourse! 1.7 i believe
    If an answer correct to 1 decimal place is required then this is correct. But if an exact answer is required then it's not correct.
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