Problem 56
Question
State the domain of the logarithmic function in interval notation. $$f(x)=\log _{2}(4 x-1)$$
Step-by-Step Solution
Verified Answer
The domain of the function is \((\frac{1}{4}, \infty)\).
1Step 1: Understand the Logarithm Function's Domain
The domain of a logarithmic function \(f(x) = \log_b(g(x))\) is determined by the condition that the argument \(g(x)\) must be greater than zero. This means we need \(g(x) > 0\).
2Step 2: Set Up the Inequality
For the given function, \(f(x) = \log_2(4x-1)\), the argument is \(4x-1\). Set up the inequality: \(4x-1 > 0\).
3Step 3: Solve the Inequality
Solve \(4x-1 > 0\) by first adding 1 to both sides to get \(4x > 1\). Then, divide each side by 4 to obtain \(x > \frac{1}{4}\).
4Step 4: Express the Solution in Interval Notation
Convert the inequality solution \(x > \frac{1}{4}\) into interval notation. The domain in interval notation is \((\frac{1}{4}, \infty)\).
Key Concepts
Domain of a FunctionInequalitiesInterval Notation
Domain of a Function
The domain of a function refers to the set of all possible input values, or "x" values, that will give an output in the function. For logarithmic functions, like any other, identifying the domain is essential.
When it comes to logarithmic functions specifically, the argument of the logarithm (what’s inside the log) must always be positive. This is because the logarithm of zero or a negative number is undefined in the real number system.
To determine the domain of a function like \( f(x) = \log_{2}(4x-1) \), we first look at the expression that follows the logarithm: \( 4x - 1 \). The condition for the domain here becomes \( 4x - 1 > 0 \). This sets the requirement that any input \( x \), when plugged into \( 4x - 1 \), must result in a value greater than zero. This is crucial as it ensures the logarithm is defined.
When it comes to logarithmic functions specifically, the argument of the logarithm (what’s inside the log) must always be positive. This is because the logarithm of zero or a negative number is undefined in the real number system.
To determine the domain of a function like \( f(x) = \log_{2}(4x-1) \), we first look at the expression that follows the logarithm: \( 4x - 1 \). The condition for the domain here becomes \( 4x - 1 > 0 \). This sets the requirement that any input \( x \), when plugged into \( 4x - 1 \), must result in a value greater than zero. This is crucial as it ensures the logarithm is defined.
Inequalities
Inequalities are mathematical expressions involving symbols such as \( >, <, \geq, \leq \) that show the relative size or order of two values. They play an important role when finding domains of functions.
In the context of our function \( f(x) = \log_{2}(4x-1) \), we use the inequality \( 4x - 1 > 0 \) to find the domain. Solving this inequality involves simple algebraic steps:
In the context of our function \( f(x) = \log_{2}(4x-1) \), we use the inequality \( 4x - 1 > 0 \) to find the domain. Solving this inequality involves simple algebraic steps:
- First, you add 1 to both sides of the inequality: \( 4x - 1 + 1 > 0 + 1 \) which simplifies to \( 4x > 1 \).
- Next, divide each side of the inequality by 4 to solve for \( x \): \( x > \frac{1}{4} \).
Interval Notation
Interval notation is a way of writing subsets of the real number line. It describes a range of values that \( x \) can take. This is often the result once an inequality is solved.
For a solution like \( x > \frac{1}{4} \), interval notation is used to express this range efficiently. When you see \((\frac{1}{4}, \infty)\), it describes all numbers greater than \( \frac{1}{4} \), up to infinity.
For a solution like \( x > \frac{1}{4} \), interval notation is used to express this range efficiently. When you see \((\frac{1}{4}, \infty)\), it describes all numbers greater than \( \frac{1}{4} \), up to infinity.
- The parentheses "( )" are used because \( x = \frac{1}{4} \) is not included (as opposed to "[ ]" which would mean inclusive).
- The infinity symbol \( \infty \) always has a parenthesis, as we can never exactly "reach" infinity.
Other exercises in this chapter
Problem 56
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