Problem 71 - Calculus Early Transcendentals

Question

Use the inverse relations between \(\ln x\) and \(e^{x}(\exp (x)),\) and the properties of \(\ln x\) to prove the following properties. a. \(\exp (0)=1\) b. \(\exp (x-y)=\frac{\exp (x)}{\exp (y)}\) c. \((\exp (x))^{p}=\exp (p x), p\) rational

Step-by-Step Solution

Answer

Question: Prove the following properties of the functions \(\ln x\) and \(\exp x\): a. \(\exp(0) = 1\) b. \(\exp(x-y) = \frac{\exp(x)}{\exp(y)}\) c. \((\exp(x))^p = \exp(px)\) for all rational numbers p. Answer: a. We proved that \(\exp(0) = 1\). b. We proved that \(\exp(x-y) = \frac{\exp(x)}{\exp(y)}\). c. We proved that \((\exp(x))^p = \exp(px)\) for all rational numbers p.

Step 1: Property (a) Proof: \(\exp(0) = 1\)

To prove that \(\exp(0) = 1\), recall that the definition of \(\exp(x)\) is as follows: \(\exp(x) = e^x\), where \(x\) is a real number. Now, we know that \(\ln(a) \cdot \ln(b) = \ln(a \cdot b)\). Using the inverse relation between \(\ln x\) and \(\exp x\), we get: \(\ln(\exp(0) \cdot \exp(0)) = \ln(\exp(0)^2)\) \(2 \cdot \ln(\exp(0)) = \ln(\exp(0)^2)\) On the other hand, we also have: \(\ln(\exp(0)) + \ln(\exp(0)) = \ln(\exp(0) \cdot \exp(0)) = \ln(\exp(0)^2)\) So, \(2 \cdot \ln(\exp(0)) = \ln(\exp(0)^2)\), and by taking the exponent, we obtain: \(\exp(2 \cdot \ln(\exp(0))) = \exp(\ln(\exp(0)^2))\) By using the inverse relation again, we get: \(e^{2 \cdot \ln(\exp(0))} = (\exp(0))^2\) Now, we set \(\exp(0) = y\), with \(y\) being a real number. Our equation becomes: \(e^{2 \cdot \ln(y)} = y^2\) From this equation, \(y\) must satisfy the property that \(y^2 = e^{2 \cdot \ln(y)}\). We can find the value of \(y\) that does this: \(y^2 = e^{2 \cdot \ln(y)}\) \(y^2 = (e^{\ln(y)})^2\) \(y^2 = y^2\) So, \(\exp(0) = y = 1\).

Step 2: Property (b) Proof: \(\exp(x-y) = \frac{\exp(x)}{\exp(y)}\)

To prove that \(\exp(x-y) = \frac{\exp(x)}{\exp(y)}\), we will use the properties of logarithms and the inverse relations between \(\ln x\) and \(\exp x\). We start by taking the logarithm of both sides: \(\ln(\exp(x-y)) = \ln \left(\frac{\exp(x)}{\exp(y)}\right)\) Now, we will use the properties of logarithms on the right side of the equation: \(\ln(\exp(x-y)) = \ln(\exp(x)) - \ln(\exp(y))\) Next, we use the inverse relation between \(\ln x\) and \(\exp x\): \(x-y = x - y\) As both sides of the equation are equal, we have proved that \(\exp(x-y) = \frac{\exp(x)}{\exp(y)}\).

Step 3: Property (c) Proof: \((\exp(x))^p = \exp(px)\) for all rational numbers p

To prove that \((\exp(x))^p = \exp(px)\) for all rational numbers p, we will again use the properties of logarithms and the inverse relations between \(\ln x\) and \(\exp x\). We start by taking the logarithm of both sides: \(\ln((\exp(x))^p) = \ln(\exp(px))\) By using the property of logarithms \(\ln(a^b) = b \cdot \ln(a)\), we get: \(p \cdot \ln(\exp(x)) = \ln(\exp(px))\) Next, we use the inverse relation between \(\ln x\) and \(\exp x\): \(p \cdot x = px\) As both sides of the equation are equal, we have proved that \((\exp(x))^p = \exp(px)\) for all rational numbers p.

Key Concepts

Natural Logarithm

The natural logarithm () is a fundamental mathematical function denoting the logarithm with the base of Euler's number, e, approximately equal to 2.71828. The natural logarithm of a positive number x is the power to which e must be raised to obtain x. In other words, (x) = y if and only if e^y = x.

One intriguing property of the natural logarithm is its relationship with exponential functions, particularly as an inverse function. This implies that for every x, (^x) = x and ^( x) = x. The natural logarithm is also continuous and differentiable, which are crucial properties allowing it to be applied across various branches of mathematics and science, such as compounding interest in finance or radioactive decay in physics.

Inverse Relations of Ln and Exp

Inverse relations describe two functions that effectively ‘undo’ each other. In the case of and the exponential function (^), they serve as perfect inverses. This means that the natural logarithm of an exponential (e^x) is x, and vice versa, the exponential of a natural logarithm e^( x) is also x.

This property is highly significant as it provides the foundation for solving exponential and logarithmic equations. For example, to isolate x in the equation e^x = y, taking the natural logarithm of both sides would yield (e^x) = y, which simplifies to x = y due to the inverse relationship.

Properties of Logarithms

Logarithms possess several useful properties that can simplify complex calculations and proofs. Among these properties are:

• Product Rule: (ab) = a + b
• Quotient Rule: (a/b) = a - b
• Power Rule: (a^b) = b a

These properties are instrumental when proving various expressions involving exponential functions as seen in our textbook exercise. By using these properties, we can transform products into sums, quotients into differences, and powers into multiples, which often makes the manipulation of equations more manageable.

Exponential Function Proofs

Proving properties of exponential functions often involves leveraging the inverse relations shared with natural logarithms as well as the properties of logarithms themselves. For instance, demonstrating that (0) = 1 can be approached by considering the power rule of logarithms and the fact that any number raised to the power of zero equals one.

Through logical reasoning and step-by-step manipulation of the equations using the aforementioned properties, we can reliably prove expressions such as (x-y) = (x)/(y) and ((x))^p = (px). Such proofs are not just mathematical exercises; they underscore the reliable predictability and structure within the world of exponential functions, which has vast applications in fields ranging from economics to engineering.