Problem 17
Question
Find the exact value of each logarithm. $$ \log \frac{1}{1000} $$
Step-by-Step Solution
Verified Answer
The exact value is -3.
1Step 1: Understand the Problem
We need to find the logarithm of \( \frac{1}{1000} \) with base 10. This means finding what power of 10 gives \( \frac{1}{1000} \).
2Step 2: Express as a Power of 10
Recognize that \( 1000 = 10^3 \). Therefore, \( \frac{1}{1000} = 10^{-3} \).
3Step 3: Apply the Logarithm Rules
Using the logarithm rule \( \log_{10}(10^x) = x \), we find \( \log_{10}(10^{-3}) = -3 \). Thus, \( \log \frac{1}{1000} = -3 \).
Key Concepts
Power of 10Logarithmic RulesBase 10 Logarithm
Power of 10
In mathematics, understanding powers of 10 is crucial, especially when working with logarithms. A power of 10 means multiplying 1 by 10 a certain number of times. For example, "10 squared" or \(10^2\) is 100 because 10 multiplied by itself equals 100.
When dealing with fractions like \(\frac{1}{1000}\), you consider it as a negative power of 10. Here, \(1000\) is \(10^3\), so \(\frac{1}{1000}\) can be expressed as \(10^{-3}\). Similarly, \(10^0\) equals 1, which is useful to remember as a quick reference.
Think of powers of 10 like the "language" of logarithms. Knowing how to convert numbers into this form simplifies solving logarithms significantly. It breaks down complex numbers into more manageable pieces that we can easily work with in further calculations.
When dealing with fractions like \(\frac{1}{1000}\), you consider it as a negative power of 10. Here, \(1000\) is \(10^3\), so \(\frac{1}{1000}\) can be expressed as \(10^{-3}\). Similarly, \(10^0\) equals 1, which is useful to remember as a quick reference.
Think of powers of 10 like the "language" of logarithms. Knowing how to convert numbers into this form simplifies solving logarithms significantly. It breaks down complex numbers into more manageable pieces that we can easily work with in further calculations.
Logarithmic Rules
Logarithmic rules are the backbone of working with logarithms, and they make seemingly complicated operations easier. One basic rule is that the logarithm of a number represents "the power" to which the base must be raised to obtain that number. For instance, in the expression \(\log_{10}(10^x) = x\), the 10 raised to the power of \(x\) gives you the original number.
Additional rules can make calculations even more straightforward:
Additional rules can make calculations even more straightforward:
- Product Rule: \(\log_b(MN) = \log_b(M) + \log_b(N)\) – Allows multiplication to turn into addition.
- Quotient Rule: \(\log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N)\) – Changes division into subtraction.
- Power Rule: \(\log_b(M^p) = p \cdot \log_b(M)\) – Simplifies dealing with exponents inside logs.
Base 10 Logarithm
The concept of a base 10 logarithm is foundational in simplifying logarithmic expressions. Often referred to as "common logarithms," these use base 10, which aligns well with our decimal system. Just think of them as the reverse of powers of 10. In practical terms, \(\log_{10}(1000)\) equals 3, because 1000 is 10 raised to the third power. Conversely, \(\log_{10}(\frac{1}{1000})\) equals \(-3\), indicating that \(10^{-3}\) produces \(\frac{1}{1000}\).
For solving exercises like finding \(\log \left(\frac{1}{1000}\right)\), it's about recognizing expressions in terms of base 10 powers. This base is widely used not only in mathematical courses but also in scientific applications due to its intuitive nature and ease of simplification. Understanding how to utilize the base 10 logarithms efficiently aids in achieving precision in various scientific and engineering problems.
For solving exercises like finding \(\log \left(\frac{1}{1000}\right)\), it's about recognizing expressions in terms of base 10 powers. This base is widely used not only in mathematical courses but also in scientific applications due to its intuitive nature and ease of simplification. Understanding how to utilize the base 10 logarithms efficiently aids in achieving precision in various scientific and engineering problems.
Other exercises in this chapter
Problem 17
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