Results 1 to 11 of 11

Math Help - Help in Calculus!!

  1. #1
    Newbie
    Joined
    Sep 2012
    From
    singapore
    Posts
    5

    Help in Calculus!!

    1. A line has equation
      In(y) = a In(x) + b








      where a = 0.796 and b = 3.538. The equation can be written in the form
      y = cx^d








      What is the value of c, to 2 decimal places?

      anyone can help wif this question?


    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Help in Calculus!!

    I assume you mean ln instead of In...as a hint, you may write:

    \ln(y)=a\ln(x)+b=\ln(x^a)+\ln(e^b)=ln(e^bx^a)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Aug 2011
    Posts
    250
    Thanks
    60

    Re: Help in Calculus!!

    Hi !

    ln(y) = a*ln(x) +b
    exp(ln(y)) = exp(a*ln(x) +b)
    exp(ln(y)) = exp(b)*exp(a*ln(x))
    simplify exp(ln(y))
    simplify exp(a*ln(x))
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2012
    From
    singapore
    Posts
    5

    Re: Help in Calculus!!

    Quote Originally Posted by MarkFL2 View Post
    I assume you mean ln instead of In...as a hint, you may write:

    \ln(y)=a\ln(x)+b=\ln(x^a)+\ln(e^b)=ln(e^bx^a)
    if that is the case, ln can be cancel off from both side which left with c = e^b?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Help in Calculus!!

    Yes, now just use the given value for b and round to two decimal places as required.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Sep 2012
    From
    singapore
    Posts
    5

    Re: Help in Calculus!!

    thanks man!! then how about ln(y) = ax + b in the form of y=c e^dx
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Help in Calculus!!

    To rewrite the right side, use u=\ln(e^u) then \ln(v)+\ln(w)=\ln(vw).
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Sep 2012
    From
    singapore
    Posts
    5

    Re: Help in Calculus!!

    Quote Originally Posted by MarkFL2 View Post
    To rewrite the right side, use u=\ln(e^u) then \ln(v)+\ln(w)=\ln(vw).
    can u explain in a simple way? i got messed up wif the v and w
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Help in Calculus!!

    Those were just meant as general variables...

    We are given:

    \ln(y)=ax+b=\ln(e^{ax})+\ln(e^b)=\ln(e^be^{ax})

    Do you see how I applied the general rules I gave above?

    Can you finish?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Sep 2012
    From
    singapore
    Posts
    5

    Re: Help in Calculus!!

    Quote Originally Posted by MarkFL2 View Post
    Those were just meant as general variables...

    We are given:

    \ln(y)=ax+b=\ln(e^{ax})+\ln(e^b)=\ln(e^be^{ax}) and solve the remaining values

    Do you see how I applied the general rules I gave above?

    Can you finish?
    same with the above... Remove both side \ln
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Help in Calculus!!

    Yes, since logarithmic functions are monotonic (one-to-one), \ln(u)=\ln(v) implies u=v.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: December 13th 2011, 10:11 PM
  2. Replies: 2
    Last Post: June 25th 2010, 11:41 PM
  3. Replies: 1
    Last Post: February 11th 2010, 08:09 AM
  4. Calculus III But doesn't require Calculus :)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 19th 2009, 05:23 PM
  5. Replies: 1
    Last Post: June 23rd 2008, 10:17 AM

Search Tags


/mathhelpforum @mathhelpforum