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Use a calculator to approximate the given logarithms to 4 decimal places. a. Avogadro's number is \(6.022 \times 10^{23}\). Approximate \(\log \left(6.022 \times 10^{23}\right)\) b. Planck's constant is \(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{sec}\). Approximate \(\log \left(6.626 \times 10^{-34}\right)\) c. Compare the value of the common logarithm to the power of 10 used in scientific notation.

Short Answer

Expert verified
a. 23.7798, b. -33.1790. The logarithms reflect the exponents of 10 used in scientific notation.

Step by step solution

01

- Understand the problem

The task is to use a calculator to find the logarithm of given scientific numbers and approximate the results to 4 decimal places.
02

- Use the logarithm properties and calculator for part (a)

To find \(\text{log}(6.022 \times 10^{23})\), break it down using the properties of logarithms: \(\text{log}(6.022) + \text{log}(10^{23})\).\(\text{log}(6.022)\) is approximately 0.7798 and \(\text{log}(10^{23})\) is 23. Using a calculator, \(\text{log} (6.022 \times 10^{23}) = 0.7798 + 23\). Therefore, \( \text{log}(6.022 \times 10^{23}) \) is approximately \( 23.7798 \).
03

- Use the logarithm properties and calculator for part (b)

To find \(\text{log}(6.626 \times 10^{-34})\), again break it down: \(\text{log}(6.626) + \text{log}(10^{-34})\). \(\text{log}(6.626)\) is approximately 0.8210 and \(\text{log}(10^{-34})\) is -34. Using a calculator, \(\text{log} (6.626 \times 10^{-34}) = 0.8210 + (-34)\). Therefore, \( \text{log}(6.626 \times 10^{-34}) \) is approximately \(-33.1790\).
04

- Compare the logarithm values to the power of 10

In part (a), the common logarithm \( \text{log}(6.022 \times 10^{23}) = 23.7798 \) shows the exponent 23 with some additional decimal values. In part (b), \( \text{log}(6.626 \times 10^{-34}) = -33.1790 \) shows the exponent -34 with additional decimal values. The values you get are close to the powers of 10 used in the scientific notation but adjusted by the logarithm of the coefficient (first part of the number).

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Key Concepts

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

Scientific Notation
Scientific notation is a way to express very large or very small numbers concisely. It uses the format: \( a \times 10^n \), where \( a \) is a number greater than or equal to 1 and less than 10 (called the coefficient), and \( n \) is an integer (the exponent). This notation helps in simplifying complex calculations and makes it easier to read and write unwieldy numbers. For example, Avogadro's number, which is roughly 602,200,000,000,000,000,000,000, is written as \( 6.022 \times 10^{23} \) in scientific notation. Similarly, very small numbers like Planck's constant, about 0.0000000000000000000000000000000006626, are written as \( 6.626 \times 10^{-34} \). This notation is particularly useful in various scientific fields, such as chemistry and physics, where extremely large or small values are common.
Properties of Logarithms
Logarithms are mathematical functions that help in solving equations involving exponents. Some useful properties of logarithms simplify calculations:
  • The product property: \( \log(ab) = \log(a) + \log(b) \). This allows us to split a product into a sum of logarithms, making it easier to handle.
  • The quotient property: \( \log \left( \frac{a}{b} \right) = \log(a) - \log(b) \). This breaks a division into a difference of logarithms.
  • The power property: \( \log(a^b) = b \cdot \log(a) \). This transforms a power into a multiplication of a logarithm.
These properties help in breaking down complex expressions into simpler forms, as seen in our example where we broke down the logarithm of a product into the sum of its parts. Understanding these properties is crucial for solving logarithmic equations efficiently.
Logarithmic Approximation
Logarithmic approximation involves using a calculator to find the logarithmic values of specific numbers to a certain number of decimal places. In the given exercise, we approximated the logarithms to 4 decimal places. For instance, \( \log(6.022) \) is approximately 0.7798 and \( \log(10^{23}) \) is exactly 23. When using the properties of logarithms, we added these values to get \( \log(6.022 \times 10^{23}) \approx 23.7798 \). Similarly, for \( \log(6.626) \approx 0.8210 \) and \( \log(10^{-34}) \approx -34 \), we get \( \log(6.626 \times 10^{-34}) \approx -33.1790 \). Logarithmic approximation is especially useful in science and engineering where precise calculations are critical.
Common Logarithm
The common logarithm, usually denoted as \( \log \), has a base of 10. It is widely used because it corresponds to the decimal numbering system. In scientific notation, the exponent indicates the power of 10, and the common logarithm of a number provides an approximate idea of that power. For instance, in the exercise, we noticed that \( \log(6.022 \times 10^{23}) \approx 23.7798 \). The integer part 23 represents the power of 10 in the scientific notation, and the decimal part 0.7798 gives additional precision. Similarly, \( \log(6.626 \times 10^{-34}) \approx -33.1790 \) where -34 is the power of 10 and -33.1790 includes the adjustment factor from the coefficient. Using common logarithms is intuitive for many scientific and engineering applications.

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Most popular questions from this chapter

Determine if the statement is true or false. For each false statement, provide a counterexample. For example, \(\log (x+y) \neq \log x+\log y\) because \(\log (2+8) \neq \log 2+\log 8\) (the left side is 1 and the right side is approximately 1.204 ). $$ \log _{8}\left(\frac{1}{w}\right)=-\log _{8} w $$

The population of Canada \(P(t)\) (in millions) since January \(1,1900,\) can be approximated by $$P(t)=\frac{55.1}{1+9.6 e^{-0.02515 t}}$$ where \(t\) is the number of years since January 1,1900 . a. Evaluate \(P(0)\) and interpret its meaning in the context of this problem. b. Use the function to predict the Canadian population on January \(1,2015 .\) Round to the nearest million. c. Use the function to predict the Canadian population on January 1,2040 . d. Determine the year during which the Canadian population will reach 45 million. e. What value will the term \(\frac{9.6}{e^{0.02515 t}}\) approach as \(t \rightarrow \infty\) ? f. Determine the limiting value of \(P(t)\).

Show that \(2\left(\frac{e^{x}-e^{-x}}{2}\right)\left(\frac{e^{x}+e^{-x}}{2}\right)=\frac{e^{2 x}-e^{-2 x}}{2}\).

(See Example 8 ) a. Estimate the value of the logarithm between two consecutive integers. For example, \(\log _{2} 7\) is between 2 and 3 because \(2^{2}<7<2^{3}\). b. Use the change-of-base formula and a calculator to approximate the logarithm to 4 decimal places. c. Check the result by using the related exponential form. $$ \log _{5} 3 $$

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