Question

Let f be the function defined by f(x)=ln(3x+2)^k for some positive constant k. If f'(2)=3, what is the value of k?

Answer 1

(A) ln3/ln8
(B) ln8
(C) 4
(D) 8
(E) 16

Answer 2

do you know how to find the derivative of
\[ \ln (3x+2)^k \]
?

Answer 3

I believe so. Is it this? \[k(\ln(3x+2))^{k-1}*\frac {3}{3x+2}\]

Answer 4

almost. d (ln u) = 1/u du
in this case u = (3x+2)^k

1/u is 1/(3x+2)^k
what is du ?

Answer 5

in other words, what is
\[ \frac{d}{dx} (3x+2)^k \]?

Answer 6

k(3x+2)^(k-1). So the derivative of ln(3x+2)^k is (k(3x+2)^(k-1))/(3x+2)?

Answer 7

still missing it. You should have a 1/u factor where u is (3x+2)^k
\[ \frac{1}{(3x+2)^k }\cdot k(3x+2)^{k-1} \cdot 3 \]

Answer 8

now simplify that, and plug in x=2

Answer 9

Ok. So it would end up as this? \[\frac{ 1 }{ (3(2)+2)^k }*k(3(2)+2)^{k-1}*3=\frac{3k(8)^{k-1}}{8^k}\]

Answer 10

ok, but it's nicer to write (3x+2)^k-1 / (3x+2)^k as 1/(3x+2) first

but using what you did, notice that 1/8^k is 8^(-k)
and 8^k * 8^-1 * 8^-k = 8^-1 or 1/8
so you have
3k/8
and they say that f'(2)=3 so
3k/8 = 3

Answer 11

I see. So now I just solve for k, which is 8, correct?

Answer 12

yes

Answer 13

Ok. Thanks for the help.

Answer 14

so this question was a bit complicated. First, you need to know how to take the derivative of ln and use the chain rule. That can be confusing without practice.

2nd, you should be comfortable with knowing how to simplify expressions with exponents
For example, my mind immediately simplified the (3x+2)^k-1 / (3x+2)^k
i.e. look for those patterns and simplify first.