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Jhannybean

  • one year ago

Practice with Logarithmic Inverse functions! For fun :) \[f(x) = \frac{e^{2x}-1}{2-e^{2x}}\]

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  1. Jhannybean
    • one year ago
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    Lmao.

  2. anonymous
    • one year ago
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    1/2 ln something.. give me a sec xD

  3. anonymous
    • one year ago
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    1/2 ln ((2x+1)/(x+1)) ?

  4. Jhannybean
    • one year ago
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    Well, you dont even need to switch the variables first before simplifying I guess.

  5. anonymous
    • one year ago
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    i'm used to it. teachers and their solutions. need to do their way if you want the mark :P

  6. Jhannybean
    • one year ago
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    \[y=\frac{e^{2x}-1}{2-e^{2x}}\]\[2y-ye^{2x} = e^{2x}-1\]\[2y+1=e^{2x}+ye^{2x}\]\[e^{2x}(y+1)=2y+1\]\[e^{2x}=\frac{2y+1}{y+1}\]\[\ln e^{2x} =\ln\left(\frac{2y+1}{y+1}\right)\]\[2x=\ln\left(\frac{2y+1}{y+1}\right)\]\[f^{-1}(x) = \ln\left[\left(\frac{2x+1}{x+1}\right)^{1/2} \right]\]

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