Question

Prove that $\ln x^a=a\ln x$for any $x>0$and any $a$by following these steps:

(a) Use Theorem $4.35$to show that $\frac{d}{dx}\left(\ln x^a\right)=\frac{a}{x}.$

(b) Compare the derivatives of $\ln x^a$ and $a\ln x$ to argue that $\ln x^a=a\ln x+C.$

(c) Use part (b) with $x=1$ and $a=1$ to show that $C=0,$ and then complete the proof.

Short answer

(part a)

(part b)

(part c)

Step-by-step solution

Introduction (part a).

1

Given Information (part a).

2

Explanation (part a).

3

Given Information (part b).

4

Explanation (part b).

5

Given Information (part c).

6

Explanation (part c).

7