/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Solve the following for \(r\). ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 21

Solve the following for \(r\). a. \(1.025=e^{r}\) b. \(\frac{1}{2}=e^{r}\) c. \(1.08=e^{r}\)

Short Answer

Expert verified
a. \( r \approx 0.0247 \). b. \( r \approx -0.693 \). c. \( r \approx 0.0770 \).

Step by step solution

01

Understand the problem

The exercise requires solving the equation for the variable \(r\) in the context of the natural exponential function \(e^r\).
02

Isolate the exponent (Part a)

For the equation \(1.025 = e^r\), take the natural logarithm (\(\text{ln}\)) on both sides to get \( \text{ln}(1.025) = \text{ln}(e^r) \).
03

Apply the natural logarithm property (Part a)

Using the property \( \text{ln}(e^r) = r \), the equation becomes \( r = \text{ln}(1.025) \).
04

Calculate the value of \(r\) (Part a)

Compute \( \text{ln}(1.025) \) to find \(r\). Using a calculator, \( r \approx 0.0247 \).
05

Isolate the exponent (Part b)

For the equation \(\frac{1}{2} = e^r\), take the natural logarithm (\(\text{ln}\)) on both sides to get \( \text{ln}(\frac{1}{2}) = \text{ln}(e^r) \).
06

Apply the natural logarithm property (Part b)

Using the property \( \text{ln}(e^r) = r \), the equation becomes \( r = \text{ln}(\frac{1}{2}) \).
07

Calculate the value of \(r\) (Part b)

Compute \( \text{ln}(\frac{1}{2}) \). Recall \( \text{ln}(\frac{1}{2}) = -\text{ln}(2) \) so, \( r \approx -0.693 \).
08

Isolate the exponent (Part c)

For the equation \(1.08 = e^r\), take the natural logarithm (\(\text{{ln}}\)) on both sides to get \( \text{ln}(1.08) = \text{ln}(e^r) \).
09

Apply the natural logarithm property (Part c)

Using the property \( \text{ln}(e^r) = r \), the equation becomes \( r = \text{ln}(1.08) \).
10

Calculate the value of \(r\) (Part c)

Compute \( \text{ln}(1.08) \) to find \(r\). Using a calculator, \( r \approx 0.0770 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Functions
Exponential functions involve a constant base raised to a variable exponent. A common exponential function is the natural exponential function, denoted as \(e^r\). Here, \(e\) is a special mathematical constant approximately equal to 2.71828. Exponential functions are fundamental in various fields, including science, finance, and engineering, due to their unique growth properties. For example, the equations in our exercise \(1.025 = e^r\), \(\frac{1}{2} = e^r\), and \(1.08 = e^r\) are all natural exponential functions where we solve for the exponent \(r\).
Solving Equations
Solving equations means finding the value of the unknown variable that makes the equation true. In our case, we need to find the value of \(r\) in the equations \(1.025 = e^r\), \(\frac{1}{2} = e^r\), and \(1.08 = e^r\). Here are the general steps:
  • First, isolate the exponential term if it's not already done.
  • Next, use logarithms to help solve for the exponent. The natural logarithm (\(\ln\)) is particularly useful when working with the base \(e\).
  • Simplify the resulting equation, and calculate the value.
For example, in the equation \(1.025 = e^r\), taking the natural logarithm on both sides helps us isolate \(r\), making it easier to solve.
Logarithmic Properties
Logarithms are the inverse operations of exponentiation. The natural logarithm, denoted by \(\ln\), specifically deals with the base \(e\). Some key properties of natural logarithms that we use in solving equations include:
  • \(\ln(e^r) = r\), because taking the natural logarithm of \(e^r\) simplifies directly to the exponent \(r\).
  • \(\ln(ab) = \ln(a) + \ln(b)\), which is useful when dealing with products inside the logarithm.
  • \(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\), which helps in cases where the logarithm involves a fraction.
In our solutions, we mainly applied the property \(\ln(e^r) = r\) to simplify and solve each equation.
Isolation of Variables
The process of isolating a variable involves manipulating an equation so that the variable of interest stands alone on one side. Here鈥檚 how it's done for an exponential equation like \(1.08 = e^r\):
  • We start by taking the natural logarithm on both sides: \(\ln(1.08) = \ln(e^r)\).
  • Next, apply the logarithm property \(\ln(e^r) = r\) to simplify the equation to \(r = \ln(1.08)\).
  • Finally, we compute the natural logarithm to find the value of \(r\).
This method quickly isolates \(r\) and helps us find its value straightforwardly. Each equation in our exercise was handled similarly to isolate and solve for \(r\).

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