Using the new definition of $\ln x$ as an integral and $e^{x}$ as the inverse of $\ln x$
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 $\frac{d}{d x}\left(\ln x^{a}\right)=\frac{a}{x}$.
(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.