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

1. Oct 4, 2012

### silvershine

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. Oct 4, 2012

### 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?

3. Oct 4, 2012

### 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.

4. Oct 4, 2012

### silvershine

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?

5. Oct 4, 2012

### Ray Vickson

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

RGV

6. Oct 4, 2012

### silvershine

It says ln[(5x-3)^k].

7. Oct 4, 2012

### Staff: Mentor

Now that that's cleared up,
f(x) = ln[(5x - 3)k] = k*ln(5x - 3)
What is f'(x)?

8. Oct 4, 2012

### silvershine

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)?

9. Oct 4, 2012

### 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.

10. Oct 4, 2012

### silvershine

Hm... So then would it be 5k/(5x-3)?

11. Oct 4, 2012

### 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.

12. Oct 4, 2012

### silvershine

Sorry.
So f'(x) would be 5k/(5x-3)?

13. Oct 4, 2012

### Staff: Mentor

Yes.

And you're given that f'(1) = 4, so k = ?

14. Oct 4, 2012

### silvershine

So then k = 8/5, right?

15. Oct 4, 2012

### Staff: Mentor

Yep.

16. Oct 4, 2012

### silvershine

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?

17. Oct 4, 2012

### 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.

18. Oct 4, 2012

### silvershine

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.

19. Oct 4, 2012

### happysauce

In some sense you are doing the product rule, but the derivative of a constant is zero so...

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