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Chapter 7: Problem 3

Let \(a>0\) and \(r \in \mathbb{Q}\). Show that \(\ln a x^{r}=\ln a+r \ln x\) for all \(x \in(0, \infty)\), assuming only that \((\ln )^{\prime} x=1 / x\) for all \(x \in(0, \infty)\).

Short Answer

Expert verified
To demonstrate that \(\ln a x^r = \ln a + r \ln x\) for all \(x \in (0, \infty)\), we differentiate both sides of the equation with respect to \(x\) using the given properties of the natural logarithm function and the chain rule. After differentiating the left side, we get \(\frac{r x^{r-1}}{x^r}\), and after differentiating the right side, we get \(r \frac{1}{x}\). Since both derivatives are equal, we can confirm that \(\ln a x^r = \ln a + r \ln x\) for all \(x \in (0, \infty)\).

Step by step solution

01

Differentiate both sides of the equation

Let's differentiate both sides of the equation \(\ln a x^r = \ln a + r \ln x\), using the fact that \((\ln )^{\prime} x=1 / x\) for all \(x \in (0, \infty)\). The chain rule tells us that if \(f(u)=\ln u\) and \(u=ax^r\), then \(\frac{d}{dx}\ln u = \frac{d(\ln u)}{du} \cdot \frac{du}{dx}\).
02

Differentiate left side of the equation

Let's start by differentiating the left side of the equation, \(\ln a x^r\). By applying the chain rule, we have: \[ \frac{d}{dx}\ln a x^r = (\ln )^{\prime} (a x^r) \cdot \frac{d}{dx}(a x^r) \] The derivative of the logarithm is given, so we have: \[ \frac{1}{a x^r} \cdot \frac{d}{dx}(a x^r) \] Now we just need to find the derivative of the product \(a x^r\). The product rule tells us that if \(u=ax\) and \(v=x^r\), then \(\frac{d}{dx}(uv) = u'v+uv'\). Therefore, we have: \[ \frac{1}{a x^r} \cdot (a \cdot r x^{r-1} + a x^r \cdot 0) \] which simplifies to: \[ \frac{r x^{r-1}}{x^r} \]
03

Differentiate right side of the equation

Now let's differentiate the right side of the equation, \(\ln a + r \ln x\). Since the derivative of a sum is the sum of the derivatives, we have: \[ \frac{d}{dx}(\ln a + r \ln x) = (\ln )^{\prime}(a) + r (\ln )^{\prime}(x) \] The derivative of the logarithm is given and we know that logarithm of a constant equal to zero, so we have: \[ 0 + r \frac{1}{x} \] which simplifies to: \[ r \frac{1}{x} \]
04

Compare the derivatives

Now we need to compare the derivatives of both sides of the equation: \[ \frac{r x^{r-1}}{x^r} = r \frac{1}{x} \] If we simplify the left side, we find that: \[ \frac{r}{x} = r \frac{1}{x} \] Since both sides are equal, this confirms that \(\ln a x^r = \ln a + r \ln x\) for all \(x \in (0, \infty)\).

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Most popular questions from this chapter

Let \(a, b \in(0, \infty)\). (i) Consider the functions \(f, g:(0, \infty) \rightarrow \mathbb{R}\) defined by \(f(x):=\log _{a} x\) and \(g(x):=\log _{b} x .\) Show that \(f\) and \(g\) have the same rate as \(x \rightarrow \infty\). (ii) Consider the functions \(f, g: \mathbb{R} \rightarrow \mathbb{R}\) defined by \(f(x):=a^{x}\) and \(g(x):=b^{x} .\) Show that the growth rate of \(f\) is less than that of \(g\) as \(x \rightarrow \infty\) if and only if \(a

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