Q. 76 - Calculus

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. 

Step-by-Step Solution

Answer

(part a)

(part b)

(part c)

Step1: Introduction (part a).

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Step2: Given Information (part a).

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Step3: Explanation (part a).

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Step4: Given Information (part b).

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Step5: Explanation (part b).

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Step6: Given Information (part c).

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Step7: Explanation (part c).

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