Chapter 13: Problem 4
Use your calculator to find the log of the following numbers. (a) \(10^{4}\) (b) \(1 \times 10^{-6}\) (c) \(1.7 \times 10^{8}\) (d) \(10^{-8}\) (e) 10
Short Answer
Expert verified
The logs of the given numbers are: (a) \(10^{4}\) is 4, (b) \(1 \times 10^{-6}\) is -6, (c) \(1.7 \times 10^{8}\) is approximately 8.23, (d) \(10^{-8}\) is -8 and (e) 10 is 1.
Step by step solution
01
Calculating the log of \(10^{4}\)
Use your calculator to calculate the logarithm to the base 10 of \(10^{4}\), which is simply 4 by the rule of logarithm for exponents (\( \log_b {b^n} = n )\).
02
Calculating the log of \(1 \times 10^{-6}\)
Use your calculator to find the log to the base 10 of \(1 \times 10^{-6}\). By the rule of logarithm for exponents, you can separate the 1 and \(10^{-6}\) as separate components and find their logs, which yields \( \log{1} + \log{10^{-6}} \). Logarithm of 1 to any base is 0 by rule, and logarithm of \(10^{-6}\) to base 10 is -6 by rule, so the answer is -6.
03
Calculating the log of \(1.7 \times 10^{8}\)
Use your calculator to find the log to the base 10 of \(1.7 \times 10^{8}\). By the rule of logarithm for products, you can separate the 1.7 and \(10^{8}\) as separate components and find their logs, which is \( \log{1.7} + \log{10^{8}} \). Using a calculator, \( \log{1.7} \) approximates 0.23. Logarithm of \(10^{8}\) to base 10 is 8 by rule, so the answer is approximately 8.23.
04
Calculating the log of \(10^{-8}\)
Use your calculator to find the log to the base 10 of \(10^{-8}\), which is simply -8 by the rule of logarithm for exponents .
05
Calculating the log of 10
Use your calculator to find the log to the base 10 of 10, which is simply 1 by the rule of logarithm for exponents .
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Logarithm Rules
Understanding logarithm rules is foundational for solving logarithm problems. A logarithm, in simplest terms, is the inverse operation to exponentiation. The notation \( \log_b(x) \) is read as "log base \( b \) of \( x \)". There are several key rules that simplify understanding and calculating logarithms.
Additionally, special cases include: \( \log_b(b) = 1 \) because \( b^1 = b \), and \( \log_b(1) = 0 \) because \( b^0 = 1 \). These rules are essential for breaking down complex logarithmic expressions.
- The product rule: \( \log_b{(xy)} = \log_b{(x)} + \log_b{(y)} \). This states that the log of a product is the sum of the logs.
- The quotient rule: \( \log_b{\left(\frac{x}{y}\right)} = \log_b{(x)} - \log_b{(y)} \). This expresses the log of a quotient as the difference of the logs.
- The power rule: \( \log_b{(x^n)} = n \log_b{(x)} \). This shows that the log of a power is the exponent times the log.
- The change of base formula: \( \log_b(x) = \frac{\log_k(x)}{\log_k(b)} \). This allows you to switch the base of the logarithm.
Additionally, special cases include: \( \log_b(b) = 1 \) because \( b^1 = b \), and \( \log_b(1) = 0 \) because \( b^0 = 1 \). These rules are essential for breaking down complex logarithmic expressions.
Logarithm Calculations
Calculating logarithms can seem tricky at first, but with practice, it becomes easier.
When approaching a logarithm problem, first identify if you can apply any basic rules. In the examples given in step-by-step solutions, numbers with a base of 10 offer direct uses of these rules.
Sometimes, you may need a calculator, like when dealing with non-integers or other bases.
For \( \log_{10}(1.7 \times 10^8) \), you should separate it into \( \log_{10}(1.7) + \log_{10}(10^8) \), calculating each part separately.
Remember to use logarithm rules as shortcuts, confirming that each calculation obeys the mathematical laws of logarithms.
When approaching a logarithm problem, first identify if you can apply any basic rules. In the examples given in step-by-step solutions, numbers with a base of 10 offer direct uses of these rules.
- For instance, \( \log_{10}(10^4) \) simplifies directly to 4 using the power rule: \( \log_b{b^n} = n \).
- Another example is \( \log_{10}(10^{-6}) \), which simplifies to -6, again using the power rule.
Sometimes, you may need a calculator, like when dealing with non-integers or other bases.
For \( \log_{10}(1.7 \times 10^8) \), you should separate it into \( \log_{10}(1.7) + \log_{10}(10^8) \), calculating each part separately.
- Typically, the first part, \( \log_{10}(1.7) \), needs a calculator, while the second part, \( \log_{10}(10^8) \), simplifies to 8.
Remember to use logarithm rules as shortcuts, confirming that each calculation obeys the mathematical laws of logarithms.
Exponents and Logarithms
Exponents and logarithms share a close relationship because logarithms are essentially inverted exponent operations.
Understanding this relationship deepens comprehension of both concepts.
By understanding this inverse relationship, it is easier to see how equations involving exponents can be converted into logarithmic form to simplify solutions.
The rule \( b^{\log_b(x)} = x \) demonstrates the interplay between the two; any number raised to the logarithm of the same base returns the original number.
This understanding is crucial since many real-world phenomena use exponential and logarithmic models, from computing interest rates to measuring earthquake magnitudes.
Understanding this relationship deepens comprehension of both concepts.
- Exponential form shows a number being multiplied by itself: e.g., \( b^n \) means \( b \) multiplied \( n \) times.
- Logarithmic form answers the question: "To what power must we raise the base \( b \) to obtain x?"
By understanding this inverse relationship, it is easier to see how equations involving exponents can be converted into logarithmic form to simplify solutions.
The rule \( b^{\log_b(x)} = x \) demonstrates the interplay between the two; any number raised to the logarithm of the same base returns the original number.
- Also, \( \log_b(b^x) = x \) confirms that logarithms are the exponents needed to achieve the number with that base.
- Using these forms interchangeably makes it easier to solve problems involving both exponents and logarithms.
This understanding is crucial since many real-world phenomena use exponential and logarithmic models, from computing interest rates to measuring earthquake magnitudes.