Domain and Range of Exponential and Logarithmic Functions

1. Exponential Functions

An exponential function is of the form: \[ f(x) = a \cdot b^x, \] where:

• \(a \neq 0\): A constant that determines the vertical stretch or reflection.
• \(b > 0\) and \(b \neq 1\): The base of the exponential.

Domain and Range

Domain: The domain of any exponential function is: \[ (-\infty, \infty), \] because the base \(b^x\) is defined for all real numbers \(x\).
Range: The range depends on \(a\):

• If \(a > 0\), the range is: \[ (0, \infty). \]
• If \(a < 0\), the range is: \[ (-\infty, 0). \]

Example 1: Exponential Growth

Consider: \[ f(x) = 2^x. \] Domain: \[ (-\infty, \infty). \] Range: \[ (0, \infty), \] because \(f(x)\) is always positive.

Example 2: Exponential Decay

Consider: \[ f(x) = -3 \cdot (0.5)^x. \] Domain: \[ (-\infty, \infty). \] Range: \[ (-\infty, 0), \] because \(f(x)\) is always negative.

2. Logarithmic Functions

A logarithmic function is of the form: \[ f(x) = a \cdot \log_b(x), \] where:

• \(a \neq 0\): A constant determining the vertical stretch or reflection.
• \(b > 0\) and \(b \neq 1\): The base of the logarithm.

Domain and Range

Domain: The domain of any logarithmic function is: \[ (0, \infty), \] because logarithms are defined only for positive inputs.
Range: The range of any logarithmic function is: \[ (-\infty, \infty), \] because logarithms can output all real numbers.

Example 3: Basic Logarithmic Function

Consider: \[ f(x) = \log_2(x). \] Domain: \[ (0, \infty). \] Range: \[ (-\infty, \infty). \]

Example 4: Reflected Logarithmic Function

Consider: \[ f(x) = -\log_5(x). \] Domain: \[ (0, \infty). \] Range: \[ (-\infty, \infty). \]

3. Applications

Example 5: Compound Interest

The formula for compound interest is: \[ A = P \cdot \left(1 + \frac{r}{n}\right)^{n \cdot t}, \] where:

• \(A\): Final amount.
• \(P\): Principal (initial investment).
• \(r\): Annual interest rate (in decimal).
• \(n\): Number of compounding periods per year.
• \(t\): Time in years.

Problem: A principal of $2000 is invested at an annual interest rate of 5%, compounded monthly. Find the amount after 10 years.

Solution:
The formula for compound interest is: \[ A = P \cdot \left(1 + \frac{r}{n}\right)^{n \cdot t}. \] Substitute the given values: \[ A = 2000 \cdot \left(1 + \frac{0.05}{12}\right)^{12 \cdot 10}. \] Step-by-step calculation:

• \(1 + \frac{0.05}{12} = 1.004167\).
• \(12 \cdot 10 = 120\).
• \((1.004167)^{120} \approx 1.647009\).
• \(A = 2000 \cdot 1.647009 \approx 3294.02\).

Final Answer: The amount is approximately $3294.02.

Example 6: Half-Life of Radioactive Decay

The formula for radioactive decay is: \[ A(t) = A_0 \cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}, \] where:

• \(A(t)\): Remaining amount at time \(t\).
• \(A_0\): Initial amount.
• \(h\): Half-life of the substance.

Problem: A radioactive isotope has a half-life of 12 years. If the initial amount is 500 g, how much remains after 36 years?

Solution: \[ A(36) = 500 \cdot \left(\frac{1}{2}\right)^{\frac{36}{12}}. \] Calculate step-by-step:

• \(\frac{36}{12} = 3\).
• \(\left(\frac{1}{2}\right)^3 = \frac{1}{8} = 0.125\).
• \(A(36) = 500 \cdot 0.125 = 62.5.\)

Final Answer: The remaining amount is 62.5 g.

4. Practice Questions

1. Find the domain and range of \(f(x) = 4^x\).
2. Find the domain and range of \(f(x) = \log_3(x)\).
3. A sum of $1000 is invested at an annual interest rate of 8%, compounded quarterly. Find the amount after 5 years.
4. The half-life of a radioactive substance is 6 years. If the initial mass is 80 g, how much remains after 24 years?
5. Solve for \(x\): \(3^x = 81\).
6. Solve for \(x\): \(\log_4(x) = 2\).
7. Explain why the range of \(f(x) = e^x\) is \((0, \infty)\).
8. Prove that the domain of \(f(x) = \ln(x)\) is \((0, \infty)\).

5. Answers

1. Domain: \((- \infty, \infty)\); Range: \((0, \infty)\).
2. Domain: \((0, \infty)\); Range: \((- \infty, \infty)\).
3. \[ A = 1000 \cdot \left(1 + \frac{0.08}{4}\right)^{4 \cdot 5} \approx 1485.95. \]
4. The formula for radioactive decay is: \[ A(t) = A_0 \cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}. \] Substituting the values: \[ A(24) = 80 \cdot \left(\frac{1}{2}\right)^{\frac{24}{6}}. \] Step-by-step calculation:
   • \(\frac{24}{6} = 4\).
   • \(\left(\frac{1}{2}\right)^4 = \frac{1}{16} = 0.0625\).
   • \(A(24) = 80 \cdot 0.0625 = 5 \, \text{g}.\)
   Final Answer: The remaining amount is: \[ A(24) = 5 \, \text{g}. \]