Hey! Sign in to get help with your study questionsNew here? Join for free to post

C3 rearranging equation

Announcements Posted on
How proud of your uni are you? Take our survey for the chance to win a holiday! 09-02-2016
  1. Offline

    ReputationRep:
    Hi,
    Stuck on ex4c q6c:

    Show that the equation e^0.8x-1/3-2x=0 can be written in the form
    x=p ln(3-2x), stating the value of p.

    Can anyone help? thanks
  2. Offline

    ReputationRep:
    Your question is not clear enough
  3. Offline

    ReputationRep:
    how?
  4. Online

    ReputationRep:
    you need to get the exponential term on the left and the other stuff on the right.
    then think how to change into a non-exponential expression on the left.
  5. Offline

    ReputationRep:
    (Original post by Shao Kahn)
    Hi,
    Stuck on ex4c q6c:

    Show that the equation e^0.8x-1/3-2x=0 can be written in the form
    x=p ln(3-2x), stating the value of p.

    Can anyone help? thanks
    You might like to add some brackets.
  6. Offline

    ReputationRep:
    (Original post by steve2005)
    You might like to add some brackets.
    you're right, should be: e^-0.8x-(1/3-2x)=0
  7. Offline

    ReputationRep:
    So its like: e^{-0.8x}-\frac13 + 2x = 0?
  8. Offline

    ReputationRep:
    (Original post by the bear)
    you need to get the exponential term on the left and the other stuff on the right.
    then think how to change into a non-exponential expression on the left.
    right, i think i'm close

    e^-0.8x-(1/3-2x)=0

    e^-0.8x=(1/3-2x)

    ln(-0.8x)=(1/3-2x)

    ln(3-2x)=1/-0.8x

    ln(3-2x)=-1.25x

    i know i'm close because in the answers p=-1.25,
    i just can't get it into the form x=p ln(3-2x)..
  9. Offline

    ReputationRep:
    is this clearer?.. e^-0.8x-(1/(3-2x))=0
  10. Online

    ReputationRep:
    (Original post by Shao Kahn)
    right, i think i'm close

    e^-0.8x-(1/3-2x)=0

    e^-0.8x=(1/3-2x)

    ln(-0.8x)=(1/3-2x)

    ln(3-2x)=1/-0.8x

    ln(3-2x)=-1.25x

    i know i'm close because in the answers p=-1.25,
    i just can't get it into the form x=p ln(3-2x)..
    taking logs is the correct way to proceed but

    ln(e-0.8x) is just -0.8x... you must also take the log of the RHS
  11. Offline

    ReputationRep:
    sorry, its definitely e^0.8x-(1/(3-2x))=0

    new to this..
  12. Offline

    ReputationRep:
    (Original post by Shao Kahn)
    sorry, its definitely e^0.8x-(1/(3-2x))=0

    new to this..
    \displaystyle e^-^{0.8x}- \frac{1}{3-2x} = 0 Finally:rolleyes:

    And you are after something in the form

    \displaystyle x = p \ln (3-2x)

    Just for you

    Is it

    \displaystyle e^-^0^.^8^x

    or

    \displaystyle e^0^.^8^x

    you are after:confused:

    Log laws

    \displaystyle a \ln (x)  = \ln (x^a)
  13. Offline

    ReputationRep:
    (Original post by Shao Kahn)
    sorry, its definitely e^0.8x-(1/(3-2x))=0

    new to this..



    Uploaded with ImageShack.us


    Is this it ?
  14. Offline

    ReputationRep:
    (Original post by Shao Kahn)
    sorry, its definitely e^0.8x-(1/(3-2x))=0

    new to this..
    Right okay so it's


    (Original post by SubAtomic)
    \displaystyle e^0^.^8^x- \frac{1}{3-2x} = 0
    (P.S. SubAtomic - to do e^{0.8x} you can just do e^{0.8x} with no need to do e^0^.^8^x)


    So, Shao Kahn. As before, take the fraction to the right hand side. Natural log both sides.

    Hint: \dfrac{1}{3-2x} = (3-2x)^{-1}
  15. Offline

    ReputationRep:
    (Original post by steve2005)



    Uploaded with ImageShack.us


    Is this it ?
    yes! but with e^-0.8x!
  16. Offline

    ReputationRep:
    (Original post by Shao Kahn)
    yes! but with e^-0.8x!
    In that case, the answer is wrong (if the answer is in fact p = -1.25)
  17. Offline

    ReputationRep:
    (Original post by Shao Kahn)
    right, i think i'm close

    e^-0.8x-(1/3-2x)=0

    e^-0.8x=(1/3-2x)

    ln(-0.8x)=(1/3-2x)

    ln(3-2x)=1/-0.8x

    ln(3-2x)=-1.25x

    i know i'm close because in the answers p=-1.25,
    i just can't get it into the form x=p ln(3-2x)..
    Here is where you went wrong.

    You should have gotten -0.8x = ln(1/3-2x) instead.
  18. Offline

    ReputationRep:
    (Original post by hassi94)
    In that case, the answer is wrong (if the answer is in fact p = -1.25)
    I think p = 1.25




    Uploaded with ImageShack.us
  19. Offline

    ReputationRep:
    (Original post by steve2005)
    I think p = 1.25




    Uploaded with ImageShack.us
    Yep that's what I got - but she said the answer says -1.25 which I was saying is wrong
  20. Offline

    ReputationRep:
    I know this post was awhile ago but:

    f(x)=e0.8x - 1/(3-2x)

    e0.8x - 1/(3-2x) = 0

    e0.8x = 1/(3-2x)

    e-0.8x = (3-2x)/1

    -0.8x = ln(3-2x)

    x= 1/(-0.8)ln(3-2x)

    therefore x=-1.25ln(3-2x)

    so P= -1.25

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide to join now Processing…

Updated: October 14, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Today on TSR

Student Money Week continues...

Find out which Q&As are happening today

Poll
How should university tuition be funded?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

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