Problem 26
Question
For the following exercises, rewrite each expression as an equivalent ratio of logs using the indicated base. $$ \log _{14}(55.875) \text { to base } 10 $$
Step-by-Step Solution
Verified Answer
\( \log_{14}(55.875) = \frac{\log_{10}(55.875)}{\log_{10}(14)} \).
1Step 1: Identify the Change of Base Formula
The change of base formula for logarithms states that for any positive numbers \( a, b, \) and base \( c \), \( \log_b{a} \) can be rewritten as \( \frac{\log_c{a}}{\log_c{b}} \). Here, we want to express \( \log_{14}(55.875) \) using base 10.
2Step 2: Apply the Change of Base Formula
Using the change of base formula, replace \( \log_{14}(55.875) \) with a ratio of two base 10 logarithms: \( \log_{14}(55.875) = \frac{\log_{10}(55.875)}{\log_{10}(14)} \). This ratio uses the common (base 10) logarithms to express the original log.
3Step 3: Verify the Calculation
Ensure that the rewritten expression \( \frac{\log_{10}(55.875)}{\log_{10}(14)} \) is equivalent to the original log expression \( \log_{14}(55.875) \) by using the properties of logarithms. This step reassures that the change of base formula is applied correctly.
Key Concepts
Understanding LogarithmsThe Role of Common LogarithmsHow to Achieve Equivalent Expressions
Understanding Logarithms
Logarithms are mathematical operations that help us solve equations where variables are in exponents. They are essentially the inverse operations of exponentiation. For example, if you have a number expressed as a power, such as \( b^x = a \), the logarithm \( \log_b{a} = x \) helps you find its exponent. This tells us how many times the base \( b \) must be multiplied by itself to reach \( a \).
Logarithms are incredibly useful in various fields, including science, engineering, and finance, because they can convert multiplicative processes into additive ones. This property helps simplify complex problems involving exponential growth or decay.
Whenever you encounter different bases, like base 14 in the original exercise, the change of base formula allows you to convert it into an easier base to work with, such as base 10, which is particularly common in calculators.
Logarithms are incredibly useful in various fields, including science, engineering, and finance, because they can convert multiplicative processes into additive ones. This property helps simplify complex problems involving exponential growth or decay.
Whenever you encounter different bases, like base 14 in the original exercise, the change of base formula allows you to convert it into an easier base to work with, such as base 10, which is particularly common in calculators.
The Role of Common Logarithms
Common logarithms are logarithms that use base 10. They are called 'common' because they are frequently used in everyday calculations. The notation \( \log_{10} \) is commonly simplified to just \( \log \).
This makes calculations more streamlined, especially when using tools like scientific calculators, which are inherently designed for base 10 logarithms. Using common logarithms, we can easily calculate or verify expressions. In our exercise, changing the base from 14 to 10 allows us to compute \( \log_{14}(55.875) \) using the more manageable base 10 logs: \( \frac{\log(55.875)}{\log(14)} \).
This conversion simplifies working with logarithms and helps provide consistent results across calculations, ensuring that even more complex logarithmic expressions can be evaluated quickly and accurately.
This makes calculations more streamlined, especially when using tools like scientific calculators, which are inherently designed for base 10 logarithms. Using common logarithms, we can easily calculate or verify expressions. In our exercise, changing the base from 14 to 10 allows us to compute \( \log_{14}(55.875) \) using the more manageable base 10 logs: \( \frac{\log(55.875)}{\log(14)} \).
This conversion simplifies working with logarithms and helps provide consistent results across calculations, ensuring that even more complex logarithmic expressions can be evaluated quickly and accurately.
How to Achieve Equivalent Expressions
An equivalent expression in mathematics is a different way of representing the same value or concept, often simplifying calculations or aligning them with current needs. The change of base formula helps us create these equivalent expressions by allowing us to convert logarithm expressions to any of our choosing.
For the exercise provided, we used the change of base formula to express \( \log_{14}(55.875) \) as \( \frac{\log_{10}(55.875)}{\log_{10}(14)} \). This equation is an equivalent expression because both forms represent the same numerical value or logical concept.
Finding equivalent expressions is crucial in problem solving because it gives us flexibility, making it easier to understand and manipulate mathematical situations. It is a powerful tool in algebra that can make challenging problems more approachable.
For the exercise provided, we used the change of base formula to express \( \log_{14}(55.875) \) as \( \frac{\log_{10}(55.875)}{\log_{10}(14)} \). This equation is an equivalent expression because both forms represent the same numerical value or logical concept.
Finding equivalent expressions is crucial in problem solving because it gives us flexibility, making it easier to understand and manipulate mathematical situations. It is a powerful tool in algebra that can make challenging problems more approachable.
Other exercises in this chapter
Problem 25
For the following exercises, determine whether the table could represent a function that is linear, exponential, or neither. If it appears to be exponential, fi
View solution Problem 25
Graph the function and its reflection about the x-axis on the same axes. $$f(x)=-4(2)^{x}+2$$
View solution Problem 26
Refer to Table 7. $$\begin{array}{|c|c|c|c|c|c|c|} \hline \boldsymbol{x} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \boldsymbol{f}(\boldsymbol{x}) & 1125 & 1495 & 2310 &
View solution Problem 26
Use a graphing calculator and this scenario: the population of a fish farm in \(t\) years is modeled by the equation \(P(t)=\frac{1000}{1+9 e^{-0.6 t}}.\) What
View solution