Turn on thread page Beta
    • Thread Starter
    Offline

    2
    ReputationRep:
    F(x) = e^x+k, x is a real number and k is a positive constant.

    Find f(ln k), simplifying your answer.

    I got e^ln k + k

    Apparently this is then k+k=2k?

    How does e^ln k turn into k?
    Offline

    13
    ReputationRep:
    (Original post by jamb97)
    F(x) = e^x+k, x is a real number and k is a positive constant.

    Find f(ln k), simplifying your answer.

    I got e^ln k + k

    Apparently this is then k+k=2k?

    How does e^ln k turn into k?
    e^{x} and \ln(x) are inverse functions, and f(f^{-1}(x))=x
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by joostan)
    e^{x} and \ln(x) are inverse functions, and f(f^{-1}(x))=x
    Sorry but that didn't really help, can you explain a bit more?
    Offline

    19
    ReputationRep:
    (Original post by jamb97)
    Sorry but that didn't really help, can you explain a bit more?
    http://www.thestudentroom.co.uk/show....php?t=3590219
    • Thread Starter
    Offline

    2
    ReputationRep:
    I think I half understand now. I just find it easier if there's a process with steps to follow.

    Say you have F(x)= e^x and F^-1(x)= ln (x).

    Then if you do F^-1(F^(x)) then you get ln e^x. From here you can use the power rule to bring x to the front of the term, and then ln (e) cancels to 1 as e^1=e, 1*x = x.

    But if you have F(F^-1(x)), you get e^ln(x). I can't see any method/process here other than simply following the rule that inverses applied to each other reverse each other. Is there any method or do you just follow that rule?
    Offline

    19
    ReputationRep:
    (Original post by jamb97)
    I think I half understand now. I just find it easier if there's a process with steps to follow.

    Say you have F(x)= e^x and F^-1(x)= ln (x).

    Then if you do F^-1(F^(x)) then you get ln e^x. From here you can use the power rule to bring x to the front of the term, and then ln (e) cancels to 1 as e^1=e, 1*x = x.

    But if you have F(F^-1(x)), you get e^ln(x). I can't see any method/process here other than simply following the rule that inverses applied to each other reverse each other. Is there any method or do you just follow that rule?
    at this stage it is more important to do rather than to deeply understand.
    learn that

    ln(ex) = x for all x
    elnx = x for all positive x


    the explanation is that

    F(F-1(x)) = F-1(F()) =x (Identity function)

    ex = exp(x)

    lnex = ln(exp(x)) =x
    elnx =exp(ln(x)) =x
    • Study Helper
    Offline

    16
    ReputationRep:
    Study Helper
    (Original post by jamb97)
    I think I half understand now. I just find it easier if there's a process with steps to follow.



    But if you have F(F^-1(x)), you get e^ln(x). I can't see any method/process here other than simply following the rule that inverses applied to each other reverse each other. Is there any method or do you just follow that rule?
    What "rule" are you expecting?

    The whole point of inverse functions is that they're defined to reverse the effect of your original function.

    So if ln x and e^x are inverse functions of each other, then e^(ln x) = x and ln(e^x) = x. This will work as long as your functions are one-to-one over a particular domain because then it makes sense to talk about finding a unique inverse value.

    (As an aside, this is the second question on this I've seen in a couple of days, and I've seen others like it before. Are teachers not teaching this concept properly now??)
    Offline

    14
    ReputationRep:
    I like to think about it like this. ln|x| = the power which e is taken to such that it equals to x. e^lnx is the same as saying e to the power of what I just said. e to the power of the number which makes e to the power of it = x will just be x if you follow logic. Sorry that i explained this quite badly, it is hard to phrase in words!
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: September 18, 2015
Poll
Is the Big Bang theory correct?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

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

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