Question

The function \(\log x\) (base \(e\) always understood) can be defined by the equation $$ \log x=\int_{1}^{x} \frac{d t}{t}, \quad x>0 . $$ a) Use this equation to evaluate \(\log 1, \log 2\), and \(\log 0.5\) approximately. b) Prove, from Eq. (a), that \(\log x\) is defined and continuous for \(00\) and \(x>0\). [Hint: Let \(F(x)=\log (a x)-\) \(\log x\). Use the result of (b) to show that \(F^{\prime}(x) \equiv 0\), so that \(F(x) \equiv\) const. Take \(x=1\) to evaluate the constant.]

Short answer

In summary, in this exercise, we have: a) Found the natural logarithm values of \(\log 1 = 0\), \(\log 2 \approx 0.6931\), and \(\log 0.5 \approx -0.6931\) using the given integral representation. b) Proved that the natural logarithm function \(\log x\) is continuous for \(0 < x < \infty\), and its derivative is \(\frac{d \log x}{dx} = \frac{1}{x}\). c) Proved the logarithmic identity \(\log(ax) = \log a + \log x\) for any \(a>0\) and \(x>0\).

Step-by-step solution

a) Evaluation of natural logarithm: \(\log 1, \log 2\) and \(\log 0.5\)

Here is the given equation: $$ \log x = \int_{1}^{x} \frac{dt}{t}, \quad x>0 $$ 1. For \(\log 1\), let \(x = 1\): $$ \log 1 = \int_{1}^{1} \frac{dt}{t} = 0 $$ 2. For \(\log 2\), let \(x = 2\) (we can only get the approximate value): $$ \log 2 = \int_{1}^{2} \frac{dt}{t} \approx 0.6931 $$ 3. For \(\log 0.5\), let \(x = 0.5\) (we can only get the approximate value): $$ \log 0.5 = \int_{1}^{0.5} \frac{dt}{t} \approx -0.6931 $$ Now, let's move to part b of the exercise.

b) Continuity and derivative of natural logarithm: \(\log x\)

1. Continuity: The integrand function \(\frac{1}{t}\) is continuous on the domain \((0, \infty)\), and thus the integral is continuous for the same domain. Therefore, the natural logarithm function \(\log x\) is continuous for \(0 < x < \infty\). 2. Differentiability: We will find the derivative of \(\log x\) with respect to \(x\). According to the Fundamental Theorem of Calculus: $$ \frac{d\log x}{dx} = \frac{d}{dx}\left(\int_{1}^{x} \frac{dt}{t}\right) = \frac{1}{x} $$ Now let's move to part c of the exercise.

c) Proving the logarithmic identity: \(\log(ax) = \log a + \log x\)

Define a function \(F(x) = \log(ax) - \log x\). We want to show that \(F'(x) \equiv 0\). Using the result from part b, we find the derivative of \(F(x)\): $$ F'(x) = \frac{d}{dx}(\log(ax) - \log(x)) = \frac{d}{dx}\log(ax) - \frac{d}{dx}\log(x) $$ Using the chain rule: $$ F'(x) = \left(\frac{1}{ax}\cdot a\right) - \frac{1}{x} = \frac{a}{ax} - \frac{1}{x}= \frac{1}{x} - \frac{1}{x} = 0 $$ So, \(F'(x) \equiv 0\), which means that \(F(x)\) is a constant function. Now, we need to find the value of this constant. Let's calculate \(F(1)\): $$ F(1) = \log(a \cdot 1) - \log(1) = \log(a) - 0 = \log(a) $$ Thus, \(F(x) \equiv \log(a)\). Therefore, we have: $$ \log(ax) = \log a + \log x $$

Key concepts

Continuous Functions

Continuous functions are an essential backbone of calculus, allowing us to analyze and predict changes in real-world systems smoothly and consistently. In this exercise, the integral form of the natural logarithm function is given by \[ \log x = \int_{1}^{x} \frac{dt}{t}, \quad x>0 \]. Since the function \(\frac{1}{t}\) under the integral is smooth and unbroken over the interval \((0, \infty)\), it means that the function \(\log x\) is continuous for \(0 < x < \infty\).

Continuous functions are critical because they ensure there are no sudden jumps or gaps in values. Such properties allow us to use calculus tools like derivatives and integrals effectively.

• Integrals of continuous functions on closed intervals are well-defined.
• Continuous functions can be approximated as closely as desired by polynomial expressions.

Understanding continuous functions aids us in predicting behavior between intervals, assuming a noticeable trend based on surrounding values.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, two core concepts of calculus. In this exercise, it plays a pivotal role in understanding the behavior of the natural logarithm function. Specifically, the theorem tells us that:

If \(F(x)\) is an antiderivative of \(f(x)\), then \(\frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x)\).

The theorem is applied as follows:

• The function \(\frac{1}{t}\) in the definite integral \(\int_{1}^{x} \frac{dt}{t}\) becomes the antiderivative \(\log x\).
• The derivative \(\frac{d}{dx}(\log x) = \frac{1}{x}\) is immediately related back to the original function within the integral.

Utilizing this theorem allows us to transition smoothly from integral equations to differential equations, ensuring solutions to integration problems can be differentiated to verify the connection between these two mathematical operations.

Logarithmic Identities

Logarithmic identities are powerful tools in simplifying complex expressions and solving logarithmic equations. One vital identity derived from the natural logarithm defined by \[ \log x = \int_{1}^{x} \frac{dt}{t}, \quad x>0 \] has been utilized in the exercise:

• \(\log(ax) = \log a + \log x\). This identity highlights how multiplying numbers can be split into a sum of logarithms.
• To prove this, consider \(F(x) = \log(ax) - \log(x)\). By differentiating and applying the results from the continuity section, we find \(F'(x) \equiv 0\), meaning \(F(x)\) is constant.

Logarithmic identities like \(\log(ax) = \log a + \log x\) are practical when dealing with exponential growth or decay, where variables naturally multiply together. These identities help to simplify solving exponential equations and analyzing products of variables in varying contexts.