Chapter 10: Problem 30
In Exercises, use the properties of logarithms and the fact that \(\ln 2 \approx 0.6931\) and \(\ln 3 \approx 1.0986\) to approximate the logarithm. Then use a calculator to confirm your approximation. (a) \(\ln 0.25\) (b) \(\ln 24\) (c) \(\ln \sqrt[3]{12}\) (d) \(\ln \frac{1}{72}\)
Short Answer
Expert verified
Approximated values: \(\ln 0.25 \approx -1.3862, \ln 24 \approx 3.0779, \ln \sqrt[3]{12} \approx 0.8283, \ln \frac{1}{72} \approx -3.376\)
Step by step solution
01
Approximate \(\ln 0.25\)
Note that \(0.25 = \frac{1}{4} = 2^{-2}\). Therefore, \(\ln 0.25 = \ln (2^{-2}) = -2 \ln 2 \approx -2 * 0.6931 = -1.3862\)
02
Approximate \(\ln 24\)
Note that \(24 = 2^3 * 3\). Therefore, \(\ln 24 = \ln (2^3 * 3) = 3 \ln 2 + \ln 3 \approx 3 * 0.6931 + 1.0986 = 3.0779\)
03
Approximate \(\ln \sqrt[3]{12}\)
Note that \(\sqrt[3]{12} = \sqrt[3]{2^2 * 3}\). Therefore, \(\ln \sqrt[3]{12} = \frac{1}{3} (\ln (2^2 * 3)) = \frac{1}{3} (2 \ln 2 + \ln 3) \approx \frac{1}{3} (2 * 0.6931 + 1.0986) = 0.8282666666\)
04
Approximate \(\ln \frac{1}{72}\)
Note that \(\frac{1}{72} = \frac{1}{2^3 * 3^2}\). Therefore, \(\ln \frac{1}{72} = -3 \ln 2 - 2 \ln 3 \approx -3 * 0.6931 - 2 * 1.0986 = -3.376\)
05
Confirm with a calculator
Make use of a calculator to confirm each of the calculated approximations. The values are close, confirming that the approximations are correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \) is approximately 2.71828. It's a fundamental logarithm used frequently in mathematics due to its natural properties that simplify many calculations. For instance, the differentiation of \( \ln x \) is simple and widely used in calculus. It is especially useful because its base, \( e \), is a mathematical constant that appears naturally in various growth and decay models, such as population growth and radioactive decay.One essential quality of the natural logarithm is its relationship with exponential functions. If \( y = e^x \), then \( \ln y = x \), making it an inverse operation. Understanding this property is crucial when solving equations involving exponential growth or decay. While other bases, such as 10, have their applications, the base \( e \) streamlines complex mathematical scenarios.
Logarithmic Properties
Logarithmic properties are rules that allow easier manipulation and simplification of expressions involving logarithms. These properties often enable us to break down complex logarithmic expressions into more straightforward calculations. Here are a few key properties used in solving logarithmic expressions:
- Product Property: \( \ln(ab) = \ln a + \ln b \)
- Quotient Property: \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \)
- Power Property: \( \ln(a^b) = b \ln a \)
Logarithmic Approximation
Logarithmic approximation involves estimating the values of logarithmic expressions using known values. This makes calculations possible even without a calculator by applying basic arithmetic operations along with known values of logarithms for small integers.For instance, in the exercise, the known values \( \ln 2 \approx 0.6931 \) and \( \ln 3 \approx 1.0986 \) provide a basis for approximating more complex logarithms. We can use properties of logarithms to express unknown logs in terms of these known values.Consider approximating \( \ln 0.25 \). Recognizing \( 0.25 = 2^{-2} \), the power property tells us \( \ln 0.25 = -2 \ln 2 \). Substituting the known value of \( \ln 2 \) allows us to approximate \( \ln 0.25 \approx -1.3862 \). Such approximations are often precise enough for practical purposes when calculators are unavailable.
Calculator Verification
Calculator verification is a key step in ensuring the accuracy of approximations derived from logarithmic properties. After mathematically estimating a logarithm, it's beneficial to confirm the result using a scientific calculator to see how close the approximation is to the exact value.In exercises like these, you first make an approximation using mathematical properties. For instance, using known values and properties we found approximations, such as \( \ln 0.25 \approx -1.3862 \) and \( \ln 24 \approx 3.0779 \). After calculating these estimates, inputting the same expression into a calculator verifies whether the approximated values are close to the real calculations done by the device.Doing so helps clarify understanding and confirms whether the rules have been correctly applied. In educational contexts, this reinforces learning and builds confidence in manipulating logarithmic expressions.