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anonymous

  • one year ago

Review: Solving Logarithmic Functions Question below.

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  1. anonymous
    • one year ago
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    Help me solve this please. \(\sf \sqrt{ln(e^2x)+12}-2=e^{ln x}\) I know that \(\sf ln e^(2x)=2x\) and \(\sf e^{ln x}=x\) So \(\sf \sqrt{2x+12} -2=x \\ \sqrt{2x+12}=x+2\) then I square on both sides to get rid off the radical: \(\sf 2x+12=(x+2)^2 \\ 2x+12=x^2+4x+4\\ 0=x^2+4x-2x+4-12 \\ 0=x^2+2x-8 \\ (x+4)(x-2) \) so x=-4 and x=2.. but how come the correct answer is only x=2? Did I do something wrong?

  2. anonymous
    • one year ago
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    causes you can't take the of a negative number

  3. anonymous
    • one year ago
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    oops should be \(\sf ln\ e^{2x}\) oh yeah i got it lol thanks

  4. anonymous
    • one year ago
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    i tried it once, had a headache for three diays

  5. anonymous
    • one year ago
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    yw

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