Problem 20
Question
Solving an Equation In Exercises 19-24, solve for \(x\) or \(b\) . $$ \begin{array}{l}{\text { (a) } \log _{3} \frac{1}{81}=x} \\ {\text { (b) } \log _{6} 36=x}\end{array} $$
Step-by-Step Solution
Verified Answer
The solutions for the equations are \( x = -4 \) for \( \log_{3} \frac{1}{81} = x \) and \( x = 2 \) for \( \log_{6} 36 = x \).
1Step 1: Understanding the basic logarithmic property
The fundamental property of logarithms states that for any positive real numbers \( b \) and \( x \), and for \( b \neq 1 \), \( \log_{b} (b^x) = x \). We will use this property to get the values of \( x \) in both the given exercises.
2Step 2: Solving for \( x \) in the equation \( \log_{3} \frac{1}{81} = x \)
Rewrite the fraction \( \frac{1}{81} \) as \( 3^{-4} \). And applying the logarithmic property, we get \( x = -4 \)
3Step 3: Solving for \( x \) in the equation \( \log_{6} 36 = x \)
Rewrite 36 as \( 6^2 \). Applying the logarithmic property, we get \( x = 2 \)
Key Concepts
Logarithmic PropertiesExponential NotationSolving Equations
Logarithmic Properties
Logarithmic properties are essential in helping us simplify and solve logarithmic equations. One key property is that the logarithm of a number, in the base of that number itself, returns the exponent to which the base must be raised to yield the original number. For example, using the property \( \log_b (b^x) = x \), we know that when you take the logarithm of a number like \( b^x \), with \( b \) as the base, you essentially retrieve \( x \).
This property is particularly useful because it allows us to transform a logarithmic equation into an easier form to understand and solve. For instance, if you encounter \( \log_3 (\frac{1}{81}) = x \), you first rewrite \( \frac{1}{81} \) as \( 3^{-4} \), taking advantage of the relation between a fraction and its exponential form.
By applying the property, \( \log_3 (3^{-4}) = -4 \), you can solve for \( x \) and see that \( x \) is \( -4 \). Recognizing these properties and applying them correctly is pivotal in efficiently solving logarithmic equations.
This property is particularly useful because it allows us to transform a logarithmic equation into an easier form to understand and solve. For instance, if you encounter \( \log_3 (\frac{1}{81}) = x \), you first rewrite \( \frac{1}{81} \) as \( 3^{-4} \), taking advantage of the relation between a fraction and its exponential form.
By applying the property, \( \log_3 (3^{-4}) = -4 \), you can solve for \( x \) and see that \( x \) is \( -4 \). Recognizing these properties and applying them correctly is pivotal in efficiently solving logarithmic equations.
Exponential Notation
Exponential notation is a way to express numbers as a base raised to the power of an exponent. This notation is incredibly useful in simplifying calculations and solving equations. For example, 81 can be rewritten as \( 3^4 \), and \( \frac{1}{81} \) becomes \( 3^{-4} \). This transformation not only makes the calculation easier but also ties directly into the properties of logarithms.
Exponential notation is powerful because it simplifies multiplication and division, especially when dealing with large numbers or fractions. It allows us to look at numbers in terms of their powers, making it easier to manipulate and work with them in equations. The reverse operation of getting us from exponential form back to its original number is what logarithms handle.
For instance, transforming 36 into \( 6^2 \) helped us solve \( \log_6 (36) = x \) easily, yielding \( x = 2 \). Using exponential notation simplifies otherwise complex logarithmic equations, especially when they inherently feature exponential relationships.
Exponential notation is powerful because it simplifies multiplication and division, especially when dealing with large numbers or fractions. It allows us to look at numbers in terms of their powers, making it easier to manipulate and work with them in equations. The reverse operation of getting us from exponential form back to its original number is what logarithms handle.
For instance, transforming 36 into \( 6^2 \) helped us solve \( \log_6 (36) = x \) easily, yielding \( x = 2 \). Using exponential notation simplifies otherwise complex logarithmic equations, especially when they inherently feature exponential relationships.
Solving Equations
Solving logarithmic equations involves a series of steps where the application of logarithmic properties and rewriting in exponential form is common. The goal is to manipulate the equation such that the logarithm is isolated, making it simple to determine the variable's value.
To solve equations like \( \log_3 \left( \frac{1}{81} \right) = x \), recognize that rewriting \( \frac{1}{81} \) as \( 3^{-4} \) is the key step. You then apply the logarithmic property \( \log_b (b^x) = x \), directly resulting in \( x = -4 \). Similarly, when dealing with \( \log_6 (36) = x \), expressing 36 as \( 6^2 \) allows us to readily determine \( x = 2 \).
Thus, the process involves:
To solve equations like \( \log_3 \left( \frac{1}{81} \right) = x \), recognize that rewriting \( \frac{1}{81} \) as \( 3^{-4} \) is the key step. You then apply the logarithmic property \( \log_b (b^x) = x \), directly resulting in \( x = -4 \). Similarly, when dealing with \( \log_6 (36) = x \), expressing 36 as \( 6^2 \) allows us to readily determine \( x = 2 \).
Thus, the process involves:
- Identifying parts of the equation that can be rewritten in exponential form.
- Using logarithmic properties to simplify and solve.
- Ensuring consistency with the base used in both exponential and logarithmic terms.
Other exercises in this chapter
Problem 20
In Exercises 17–22, find the limit. $$ \lim _{x \rightarrow-\infty} \operatorname{csch} x $$
View solution Problem 20
Finding an Indefinite Integral In Exercises \(1-20\) , find the indefinite integral. $$ \int \frac{x-2}{(x+1)^{2}+4} d x $$
View solution Problem 20
Sketching a Graph In Exercises \(17-22,\) sketch the graph of the function. $$ y=e^{x-1} $$
View solution Problem 20
In Exercises 19–28, use the properties of logarithms to expand the logarithmic expression. $$ \ln \sqrt{x^{5}} $$
View solution