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Find k in f(x) = ln(5x-3)^k, when f'(1) = k?

  1. Oct 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the value of the constant, k, when f(x) = ln(5x-3)^k and f'(1) = 4


    2. Relevant equations
    f(x) = ln(5x-3)^k
    f'(1) = 4


    3. The attempt at a solution
    I've tried:
    f'(x) = [kln(5x-3)^(k-1)] [1/(5x-3)] (5)
    And when I simplified that, I got
    f'(x) = 5kln(5x-3)^(k-1) / (5x - 3)

    I tried plugging in 4 and 1 so that
    4 = 5kln(2)^(k-1) / 2
    8 = 5kln(2)^(k-1)
    8/5 = kln(2)^(k-1)

    And now I'm stumped.
     
    Last edited by a moderator: Oct 4, 2012
  2. jcsd
  3. Oct 4, 2012 #2

    Mark44

    Staff: Mentor

    It seems to me that you're doing a lot of extra work that you don't need to do.

    ln(5x - 3)k = k*ln(5x - 3)

    Or did you mean [ln(5x - 3)]k?
     
  4. Oct 4, 2012 #3

    Mark44

    Staff: Mentor

    BTW, this is obviously a calculus problem, so should not have been posted in the precalc section. I am moving it to the Calculus & Beyond section.
     
  5. Oct 4, 2012 #4
    Sorry about posting it in the wrong section! This was a bonus question in my precalc homework.

    From the looks of it, the question says ln(5x - 3)^k.

    So would I do:
    4 = kln(2)
    4/k = ln(2)
    k = ln(2)/4?
     
  6. Oct 4, 2012 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    To repeat Mark44's question: do you mean [ln(5x-3)]^k or ln[(5x-3)^k]? It makes a big difference.

    RGV
     
  7. Oct 4, 2012 #6
    It says ln[(5x-3)^k].
     
  8. Oct 4, 2012 #7

    Mark44

    Staff: Mentor

    Now that that's cleared up,
    f(x) = ln[(5x - 3)k] = k*ln(5x - 3)
    What is f'(x)?
     
  9. Oct 4, 2012 #8
    f'(x) would k[1/(5x+3)](5) + ln(5x-3), with the chain rule, right? Which would simplify to 5k/(5x-3) + ln(5x -3)?
     
  10. Oct 4, 2012 #9

    Mark44

    Staff: Mentor

    No. You should get only one term. I think you are using the product rule incorrectly. There really is no need to use the product rule when one factor is a constant.
     
  11. Oct 4, 2012 #10
    Hm... So then would it be 5k/(5x-3)?
     
  12. Oct 4, 2012 #11

    Mark44

    Staff: Mentor

    Who is "it"? Are you talking about f(x) or f'(x)?

    I actually know what you're referring to, but I'm trying to get you to be more precise in what you're writing.
     
  13. Oct 4, 2012 #12
    Sorry.
    So f'(x) would be 5k/(5x-3)?
     
  14. Oct 4, 2012 #13

    Mark44

    Staff: Mentor

    Yes.

    And you're given that f'(1) = 4, so k = ?
     
  15. Oct 4, 2012 #14
    So then k = 8/5, right?
     
  16. Oct 4, 2012 #15

    Mark44

    Staff: Mentor

    Yep.
     
  17. Oct 4, 2012 #16
    Alright. So to clarify:
    When you have a constant times a function, when you take the derivative, you don't have to use the product rule?
     
  18. Oct 4, 2012 #17

    Mark44

    Staff: Mentor

    I'm hopeful that a couple of points have sunk in.

    1. When you have ln[<something>k], this is equal to k*ln[<something>], as long as <something> is greater than 0.
    If you have to differentiate a function like this, it is much easier to differentiate the latter form.

    2. Never use the product rule to differentiate a product where one factor is a constant. Instead, use the constant multiple rule. Although there is nothing wrong with using the product rule in this case, the product rule is more complicated to use, so offers a greater chance of introducing a mistake.

    A corollary to 2 is
    3. Never use the quotient rule if the denominator is a constant. For example, f(x) = ##\frac{\sqrt{x}}{3}##. Write this as ##\frac{1}{3} \sqrt{x} ##
    Pull the constant out as 1/k and then use the constant multiple rule. Again, it's not wrong to use the quotient rule, but it's even more complicated than the product rule, so the chances of making a mistake are even greater.
     
  19. Oct 4, 2012 #18
    They have! Or they will, once I get more practice with them.

    Thanks for walking me through! I can do the rest of these bonus problems on my own now.
     
  20. Oct 4, 2012 #19
    In some sense you are doing the product rule, but the derivative of a constant is zero so...
     
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