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Math Help - How do i solve this problem?

  1. #1
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    How do i solve this problem?



    Do i do : lnx^2-ln(1-x) = lnx + ln2x - ln (1+x)?

    if so... what is my next step?
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  2. #2
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    \ln\left(\frac{x^2}{1-x}\right) = \ln\left(\frac{2x^2}{1+x}\right)<br />

    \frac{x^2}{1-x} = \frac{2x^2}{1+x}
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  3. #3
    Member javax's Avatar
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    Quote Originally Posted by Diggidy View Post


    Do i do : lnx^2-ln(1-x) = lnx + ln2x - ln (1+x)?

    if so... what is my next step?
    find x?
    if so here we go...
    \ln{\left(\frac{x^2}{1-x}\right)} = \ln{\left(\frac{x 2x}{1+x}\right)}<br />
    \Rightarrow
    \frac{x^2}{1-x}=\frac{2x^2}{1+x}

    now cross multiply and you'll find x
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  4. #4
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    got it, thank you
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