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

[silvershine]

Homework Statement

Find the value of the constant, k, when f(x) = ln(5x-3)^k and f'(1) = 4

Homework Equations

f(x) = ln(5x-3)^k
f'(1) = 4

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.

 

[Mark44]

Homework Statement

Find the value of the constant, k, when f(x) = ln(5x-3)^k and f'(1) = 4

Homework Equations

f(x) = ln(5x-3)^k
f'(1) = 4

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.

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?

 

[Mark44]

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.

 

[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?

 

[Ray Vickson]

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?

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

RGV

 

[silvershine]

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

RGV

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

 

[Mark44]

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

 

[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)?

 

[Mark44]

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)?

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.

 

[silvershine]

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.

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

 

[Mark44]

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

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.

 

[silvershine]

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

 

[Mark44]

Yes.

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

 

[silvershine]

So then k = 8/5, right?

 

[Mark44]

Yep.

 

[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?

 

[Mark44]

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.

 

[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.

 

[happysauce]

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?

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