Problem 19
Question
In this Study Set, assume that all variables represent positive numbers and \(b \neq 1\). Evaluate each expression. See Example \(1 .\) $$ \log _{6} 1 $$
Step-by-Step Solution
Verified Answer
\( \log_6 1 = 0 \) because any number raised to the power of 0 is 1.
1Step 1: Recall the Property of Logarithms
The basic property of logarithms states that \( \log_b a = c \) if and only if \( b^c = a \). In this exercise, we need to evaluate \( \log_6 1 \) and find its equivalent power form \( 6^c = 1 \).
2Step 2: Simplify the Expression Using Logarithmic Property
The logarithmic property \( \log_b 1 = 0 \) for any base \( b eq 1 \) comes from the fact that any non-zero number raised to the power of 0 equals 1, i.e., \( b^0 = 1 \).
3Step 3: Apply the Logarithmic Property
Using the property from Step 2, evaluate \( \log_6 1 \). Since \( 6^0 = 1 \), it follows directly that \( \log_6 1 = 0 \).
Key Concepts
Properties of LogarithmsLogarithmic ExpressionsExponents
Properties of Logarithms
Logarithms have unique properties that simplify expressions and make it easier to solve various mathematical problems. One vital property involves understanding how logs relate exponential expressions.
The key principle is that a logarithm answers the question: "To what power must the base be raised to get a certain number?"
For instance:
The key principle is that a logarithm answers the question: "To what power must the base be raised to get a certain number?"
For instance:
- Property: If \( \log_b a = c \), then equivalently, \( b^c = a \).
- Another essential property: \( \log_b 1 = 0 \). This is because any non-zero number \( b \) raised to the power of zero equals 1, thereby the expression \( b^0 = 1 \) holds true for any base \( b eq 1 \).
Logarithmic Expressions
Logarithmic expressions involve using logarithms to transform and evaluate mathematical problems. Often, they appear in forms where you're asked to simplify or calculate specific log values.
Understanding how to evaluate these expressions involves knowing the base of the logarithm and the number inside the log.
Key strategies include:
Understanding how to evaluate these expressions involves knowing the base of the logarithm and the number inside the log.
Key strategies include:
- Identifying the base and evaluating powers. For instance, with \( \log_6 1 \), recognize that you're trying to find what power 6 must be raised to be equal to 1.
- Using properties, such as \( \log_b 1 = 0 \), which simplifies many complex expressions quickly.
- Transforming an expression using properties like the power rule and product rule to make them more manageable.
Exponents
Exponents are closely linked to logarithms. They represent repeated multiplication and help describe the relationship between numbers in an exponential format.
When you explore logarithms, you encounter alongside how exponents work because:
When you explore logarithms, you encounter alongside how exponents work because:
- An equation like \( b^c = a \) is linked to \( \log_b a = c \) wherein the exponent \( c \) makes it clear "how many times" the base \( b \) multiplies itself to reach \( a \).
- Zero exponents play a crucial role due to properties like \( b^0 = 1 \), explaining why \( \log_b 1 = 0 \) holds for any base \( b eq 1 \).
- Understanding negative and fractional exponents can further expand understanding, especially since they translate into roots of a number and reciprocal multiplication respectively.
Other exercises in this chapter
Problem 18
Explain the difference in the meaning of the \(-1\) in the notation \(f^{-1}(x)\) as compared with \(a^{-1}\)
View solution Problem 19
Let \(f(x)=2 x+1\) and \(g(x)=x-3 .\) Find each function and give its domain. See Example 1. $$ g / f $$
View solution Problem 19
Complete each solution. Solve: \(\quad 2^{x}=7\) $$ \begin{aligned} _2^{x} &=\log 7 \\ x _=\log 7 \\ x &=\frac{\log 7}{\log 2} \\ x & \approx \end{aligned} $$
View solution Problem 19
a. Use a calculator to complete the table of values for \(f(x)=\log x .\) Round to the nearest hundredth. b. Graph \(f(x)=\log x .\) Note that the units on the
View solution