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Math Help - logs

  1. #1
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    logs

    how do i solve this equation?

    log3(x^2)-log3(3x+2)=0

    if i cube the logs does that undo them?
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  2. #2
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    Quote Originally Posted by teddyryan View Post
    how do i solve this equation?

    log3(x^2)-log3(3x+2)=0

    if i cube the logs does that undo them?
    Rewrite the equation:

    \log_3(x^2)-\log_3(3x+2)=0~\implies~\log_3\left(\dfrac{x^2}{3x  +2}  \right)=0\ ,\ x > -\frac23

    Now use the LHS and the RHS as exponents to the base 3:

    \dfrac{x^2}{3x+2} = 1~\implies~x^2-3x-2=0

    This is a quadratic equation. Use the quadratic formula to solve this equation. You'll get 2 solutions, but be careful: One isn't valid. Why?
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  3. #3
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    it wont be valid if its negative correct? logarithms have a vertical asymptote at 0
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  4. #4
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    Quote Originally Posted by teddyryan View Post
    it wont be valid if its negative correct? logarithms have a vertical asymptote at 0
    Correct on the negative part but it's a horizontal asymptote at x=0
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  5. #5
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    Uhh, no, the y-axis is vertical!
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