Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Chena804

  • 4 years ago

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

  • This Question is Closed
  1. lalaly
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    use \[\ln(\frac{a}{b})=lna-lnb\]

  2. lgbasallote
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    then use \(\large \log_a b = c \rightarrow a^c = b\)

  3. lalaly
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    easier way take e of both sides since its ln

  4. lgbasallote
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    as the queen orders... \(\ln a = x \rightarrow e^x = a\)

  5. lalaly
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    hehe try it @Chena804

  6. miat
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    integration?

  7. thunderemperor
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\ln x-\ln (x-1)=1\] \[\ln (x \div(x-1))=1\] \[\left(\begin{matrix}x\\ x-1\end{matrix}\right)=e ^{1}\] \[x=(x-1)e ^{1}\] You can solve the rest. ;)

  8. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy