Hey there! Sign in to join this conversationNew here? Join for free

Does e^ln = 1? Watch

Announcements
    • Thread Starter
    Offline

    2
    ReputationRep:
    Does e^ln = 1?
    Any help would be appreciated!!
    • TSR Support Team
    • Clearing and Applications Advisor
    Offline

    21
    ReputationRep:
    (Original post by metaljoe)
    Does e^ln = 1?
    Any help would be appreciated!!
    e to the ln of what? It's true that e^{ln|x|}=|x| but that's only equal to 1 for x=1.
    • Very Important Poster
    Offline

    21
    ReputationRep:
    (Original post by metaljoe)
    Does e^ln = 1?
    Any help would be appreciated!!
    e^ln is an invalid expression - ln is not taking in any arguments.
    Offline

    18
    ReputationRep:
    Yes.
    Offline

    13
    ReputationRep:
    e^Ln(1) = 1
    e^ln doesn't mean anything, you can't have a ln without a value of some kind
    Offline

    3
    ReputationRep:
    ln(1) = 0, e^0 = 1 so e^l^n^(^1^)  = 1
    • Thread Starter
    Offline

    2
    ReputationRep:
    The question I was answering was:

    Solve the following equations, giving your answers in exact form:

    ln | 3x - 1 | = 4

    I took both sides to the power of e leaving:

    e^ln|3x-1| = e^4

    So I was wondering if e^ln = 1 times what ever is the power of e^ln?
    • Community Assistant
    Offline

    18
    ReputationRep:
    (Original post by metaljoe)
    The question I was answering was:

    Solve the following equations, giving your answers in exact form:

    ln | 3x - 1 | = 4

    I took both sides to the power of e leaving:

    e^ln|3x-1| = e^4

    So I was wondering if e^ln = 1 times what ever is the power of e^ln?
    You have to have the natural log of something.  e^x and  ln(x) are inverses and  f^{-1}(fg(x)) = g(x)

    so if f(x) = e^x, then  f^{-1}(x) =ln(x)

    so, e^ln|3x-1| = |3x-1|
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by NotNotBatman)
    You have to have the natural log of something.  e^x and  ln(x) are inverses and  f^{-1}(fg(x)) = g(x)

    so if f(x) = e^x, then  f^{-1}(x) =ln(x)

    so, e^ln|3x-1| = |3x-1|
    Cheers buddy!!!
    • Thread Starter
    Offline

    2
    ReputationRep:
    Thanks everyone all sorted now!!!!
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Should Spain allow Catalonia to declare independence?
    Help with your A-levels

    All the essentials

    The adventure begins mug

    Student life: what to expect

    What it's really like going to uni

    Rosette

    Essay expert

    Learn to write like a pro with our ultimate essay guide.

    Uni match

    Uni match

    Our tool will help you find the perfect course for you

    Study planner

    Create a study plan

    Get your head around what you need to do and when with the study planner tool.

    Study planner

    Resources by subject

    Everything from mind maps to class notes.

    Hands typing

    Degrees without fees

    Discover more about degree-level apprenticeships.

    A student doing homework

    Study tips from A* students

    Students who got top grades in their A-levels share their secrets

    Study help links and info

    Can you help? Study help unanswered threadsRules and posting guidelines

    Sponsored content:

    HEAR

    HEAR

    Find out how a Higher Education Achievement Report can help you prove your achievements.

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.