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Chena804 Group Title

Solve. Please show steps. lnx-ln(x-1)=1

  • 2 years ago
  • 2 years ago

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  1. lalaly Group Title
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    use \[\ln(\frac{a}{b})=lna-lnb\]

    • 2 years ago
  2. lgbasallote Group Title
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    then use \(\large \log_a b = c \rightarrow a^c = b\)

    • 2 years ago
  3. lalaly Group Title
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    easier way take e of both sides since its ln

    • 2 years ago
  4. lgbasallote Group Title
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    as the queen orders... \(\ln a = x \rightarrow e^x = a\)

    • 2 years ago
  5. lalaly Group Title
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    hehe try it @Chena804

    • 2 years ago
  6. miat Group Title
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    integration?

    • 2 years ago
  7. thunderemperor Group Title
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    \[\ln x-\ln (x-1)=1\] \[\ln (x \div(x-1))=1\] \[\left(\begin{matrix}x\\ x-1\end{matrix}\right)=e ^{1}\] \[x=(x-1)e ^{1}\] You can solve the rest. ;)

    • 2 years ago
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