Problem 26
Question
For the following exercises, rewrite each expression as an equivalent ratio of logs using the indicated base. \(\log _{14}(55.875)\) to base 10
Step-by-Step Solution
Verified Answer
\(\log_{14}(55.875) = \frac{\log_{10}(55.875)}{\log_{10}(14)}\) using base 10 logs.
1Step 1: Identify the Given Logarithm
We are given the logarithm \(\log_{14}(55.875)\) and asked to rewrite it in terms of a ratio of logarithms with base 10.
2Step 2: Use the Change of Base Formula
The change of base formula for logarithms is \( \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \), where \(c\) is the new base. Here, we want to express the log using base 10, so we set \(c = 10\).
3Step 3: Rewrite Using Base 10
According to the change of base formula: \( \log_{14}(55.875) = \frac{\log_{10}(55.875)}{\log_{10}(14)} \). This expresses \(\log_{14}(55.875)\) as a ratio of base 10 logs.
Key Concepts
Logarithmic expressionsBase 10 logarithmsRatio of logsLogarithmic conversion
Logarithmic expressions
Logarithmic expressions form the foundation for understanding various phenomena, especially in scientific and mathematical contexts. Essentially, a logarithm answers the question: "To what power must the base be raised, to produce a given number?"
Think of it as the reverse operation of exponentiation. If you have an equation like \( b^x = a \), the logarithmic form would be \( x = \log_b(a) \). Here, "b" is the base, "a" is the number you're examining, and "x" is the exponent. This is why the expression \( \log_b(a) \) gives the power that b must be raised to, so that it equals a.
When working with logarithmic expressions, it's helpful to familiarize yourself with properties such as:
Think of it as the reverse operation of exponentiation. If you have an equation like \( b^x = a \), the logarithmic form would be \( x = \log_b(a) \). Here, "b" is the base, "a" is the number you're examining, and "x" is the exponent. This is why the expression \( \log_b(a) \) gives the power that b must be raised to, so that it equals a.
When working with logarithmic expressions, it's helpful to familiarize yourself with properties such as:
- Product Rule: \( \log_b(xy) = \log_b(x) + \log_b(y) \)
- Quotient Rule: \( \log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y) \)
- Power Rule: \( \log_b(x^n) = n \cdot \log_b(x) \)
Base 10 logarithms
Base 10 logarithms are also known as common logarithms. They're widely used in real-world applications, including science and engineering, because they relate closely to metric scales that are also base 10, like pH levels or the Richter scale for earthquakes.
When you see \( \log(x) \) without a specified base, it's typically understood to be base 10. This simplifies notation and handling because you don’t have to specify the base. In your calculator or computer, the "log" button usually operates as a base 10 logarithm.
Using the properties of logarithms with base 10 offers a streamlined way of solving problems. Here are a few practical points:
When you see \( \log(x) \) without a specified base, it's typically understood to be base 10. This simplifies notation and handling because you don’t have to specify the base. In your calculator or computer, the "log" button usually operates as a base 10 logarithm.
Using the properties of logarithms with base 10 offers a streamlined way of solving problems. Here are a few practical points:
- Base 10 logarithms can help convert multiplication into addition, simplifying arithmetic operations.
- They're helpful for expressing very large or small numbers in a manageable way (like expressing "thousands" or "millionths").
- Understanding base 10 logs are crucial in scientific notation and for many scientific calculations.
Ratio of logs
The concept of the ratio of logs arises frequently when multiplying or dividing numbers represented as logarithms. This concept is quite powerful in simplifying complex computations involving different logarithmic bases.
The change of base formula, \( \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \), which transforms any logarithm into a ratio of logs based on a new base c, is central to this topic. This transformation is particularly significant when you are limited to using base 10 logs, as most calculators readily provide these.
When you see a ratio of logs, you're effectively converting the problem into this form using another base, often for convenience. For example:
The change of base formula, \( \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \), which transforms any logarithm into a ratio of logs based on a new base c, is central to this topic. This transformation is particularly significant when you are limited to using base 10 logs, as most calculators readily provide these.
When you see a ratio of logs, you're effectively converting the problem into this form using another base, often for convenience. For example:
- This conversion is crucial when solving exponential equations where directly calculating in one base isn't possible or practical.
- In cases where exact calculation might be required, such as in scientific data analysis, using a ratio of logs ensures accuracy by circumventing rounding errors inherent in manual computations.
Logarithmic conversion
Logarithmic conversion involves changing a log from one base to another base. It's an essential operation, especially when dealing with different technological or scientific tools that may not support arbitrary bases natively.
This conversion is executed using the change of base formula: \( \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \). By applying this formula, you can express a logarithm in whatever base is most convenient or required for your purposes. For instance:
If you need to calculate \( \log_2(16) \) but only have a tool that uses base 10 logs, convert it by using the formula:
This conversion is executed using the change of base formula: \( \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \). By applying this formula, you can express a logarithm in whatever base is most convenient or required for your purposes. For instance:
If you need to calculate \( \log_2(16) \) but only have a tool that uses base 10 logs, convert it by using the formula:
- Set the new base \( c \) (often 10) to work within calculator restrictions.
- Apply: \( \log_2(16) = \frac{\log_{10}(16)}{\log_{10}(2)} \), simplifying the calculations.
Other exercises in this chapter
Problem 26
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