Problem 8
Question
For the following exercises, state the domain and range of the function. \(g(x)=\log _{5}(2 x+9)-2\)
Step-by-Step Solution
Verified Answer
Domain: \((\frac{-9}{2}, \infty)\); Range: \((-\infty, \infty)\).
1Step 1: Identify the Type of Function
The function given is a logarithmic function of the form \(g(x) = \log_{5}(2x+9) - 2\). We need to determine the domain and range of this function.
2Step 2: Determine the Domain
For a logarithmic function \(\log_{b}(f(x))\) to be defined, the argument \(f(x)\) must be greater than zero. Here \(f(x) = 2x + 9\). Therefore, we set up the inequality:\[2x + 9 > 0\]Solve for \(x\):\[2x > -9\]\[x > -\frac{9}{2}\]The domain of \(g(x)\) is all real numbers \(x\) such that \(x > -\frac{9}{2}\).
3Step 3: Determine the Range
The range of a logarithmic function \(\log_{b}(f(x))\) is all real numbers. Therefore, the range of \(\log_{5}(2x+9)\) without the subtraction of 2 is all real numbers. The subtraction of 2 shifts the entire graph down by 2 units, but the range is still all real numbers:\[\text{Range of \(g(x)\): } (-\infty, \infty)\]
4Step 4: State the Domain and Range
The domain and range are:- Domain: \((\frac{-9}{2}, \infty)\), which represents all \(x\) such that \(x > \frac{-9}{2}\).- Range: \((-\infty, \infty)\), which includes all possible values for \(g(x)\).
Key Concepts
Domain of a FunctionRange of a FunctionInequalities in Mathematics
Domain of a Function
The domain of a function refers to all the possible input values (often represented by \(x\)) for which the function is defined. For a function like \(g(x) = \log_{5}(2x+9) - 2\), you must consider the nature of the logarithmic expression to determine its domain.
For logarithmic functions, the expression inside the logarithm must always be positive. This is because the logarithm of a non-positive number is undefined in the real number system.
For logarithmic functions, the expression inside the logarithm must always be positive. This is because the logarithm of a non-positive number is undefined in the real number system.
- Given \(g(x) = \log_{5}(2x+9) - 2\), the term \(2x + 9\) must be greater than zero.
- Set up the inequality \(2x + 9 > 0\) and solve for \(x\).
- This simplifies to \(x > -\frac{9}{2}\), indicating that the domain comprises all real numbers greater than \(-\frac{9}{2}\).
Range of a Function
The range of a function encompasses all possible output values (often represented by \(y\) or \(g(x)\)) that the function can produce. Understanding the behavior of logarithmic functions is crucial here.
Logarithmic functions typically possess a range of all real numbers. When an adjustment, such as a vertical shift, is made to the function, the range still spans all real numbers. In the case of \(g(x) = \log_{5}(2x+9) - 2\), the \(-2\) indicates a downward shift by 2 units.
Logarithmic functions typically possess a range of all real numbers. When an adjustment, such as a vertical shift, is made to the function, the range still spans all real numbers. In the case of \(g(x) = \log_{5}(2x+9) - 2\), the \(-2\) indicates a downward shift by 2 units.
- Despite this shift, the span of possible output still remains all real numbers.
- The output \(g(x)\) of this function can still take any real number value.
Inequalities in Mathematics
Inequalities are mathematical expressions involving the symbols \(>\), \(<\), \(\geq\), or \(\leq\), used to describe the relative size of two values. They are essential when determining domains for certain functions, such as logarithmic ones.
For a function like \(g(x) = \log_{5}(2x+9) - 2\), the inequality \(2x + 9 > 0\) helps establish the domain. Solving this inequality requires isolating \(x\):
For a function like \(g(x) = \log_{5}(2x+9) - 2\), the inequality \(2x + 9 > 0\) helps establish the domain. Solving this inequality requires isolating \(x\):
- Start by subtracting 9 from each side: \(2x > -9\).
- Then, divide every term by 2 to solve for \(x\): \(x > -\frac{9}{2}\).
- This translates to the domain: \(x\) must be greater than \(-\frac{9}{2}\).
Other exercises in this chapter
Problem 8
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