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 \(\frac{d}{d x}\left(\ln x^{a}\right)=\frac{a}{x}\).

(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.

Short answer

(a).

$\ln y=a+b$

(b).

$\ln z=a+b$

(c).

$e^{a+b}=e^a{e}^b$

Step-by-step solution

Part (a) Step 1: Given information

Given expression is

$e^{a+b}=e^a{e}^b$

Part (a) Step 2: Simplification

Let,

$y=e^{a+b}$

Then,

$$\begin{gathered} y=e^{u+b} \\ \ln y=\ln e^{a+b}[\text{Taking ln on both sides]} \\ \ln y=a+b\left[\ln e^1=1\right] \end{gathered}$$

Thereforelocalid="1668419670368" $\ln y=a+b$.

Part (b) Step 1: Given information

Given expression is

$e^{a+b}=e^a{e}^b$

Part (b) Step 2: Simplification

Let,

$z=e^a{e}^b$

Then.

$$\begin{gathered} z=e^a{e}^b \\ \ln z=\ln \left(e^a{e}^b\right)[\text{Taking}\ln \text{on both sides]} \\ \ln z=\ln e^a+e^b[\text{property of logarithm]} \\ \ln z=a+b\left[\ln e^1=1\right] \end{gathered}$$

Therefore, $\ln z=a+b$.

Part (c) Step 1: Given information

Given expression is

$e^{a+b}=e^a{e}^b$

Part (c) Step 2: Simplification

The results of part a and part b are.

$$\begin{gathered} \ln y=a+b \\ \ln z=a+b \end{gathered}$$

So, $\ln z=\ln yy=z$

Therefore, $e^{a+b}=e^a{e}^b$.