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Math Help - solution to equation with logs

  1. #1
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    solution to equation with logs

    I want to know why log(9x^2) = 0 has two solutions. If we write it as log(9) + log(x^2) = 0, ==> log(9) + 2 log(x) = 0 ==> 2log(x) = -log(9) ==> log(x) =-[log(9)]/2 ==> x = 10 ^ ([-log(9)]/2), there is nothing in the algebraic manipulation that would suggest two solutions because each move was via some rule of common logs. Of course, i understand intuitively why it has two solutions, but why does the algebraic manipulation allow us to arrive at only one solution?
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  2. #2
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    Quote Originally Posted by eniuqvw View Post
    I want to know why log(9x^2) = 0 has two solutions. If we write it as log(9) + log(x^2) = 0, ==> log(9) + 2 log(x) = 0 ==> 2log(x) = -log(9) ==> log(x) =-[log(9)]/2 ==> x = 10 ^ ([-log(9)]/2), there is nothing in the algebraic manipulation that would suggest two solutions because each move was via some rule of common logs. Of course, i understand intuitively why it has two solutions, but why does the algebraic manipulation allow us to arrive at only one solution?
    \log x^2 \neq 2 \log x.

    \log x^2 = 2 \log {\color{red}|} x{\color{red}|}. Two solutions follow from this.

    The easier approach to the question is to solve 9 x^2 = 1.
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  3. #3
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    oops

    ok, i should have realized that the rule log(b^t) = t [log(b)] is defined for positive b, whereas in log(9x^2), x can take any real value. Thanks for the help
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