Exponential Growth

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Answer Key & Explanations

Exponential Growth · Grade 9 · Worksheet 1

1. 2^(x+3) - 2^(x+1) = 96 Answer: 4 Solution: Factor out 2^(x+1) from both terms: 2^(x+1)(2^2 - 1) = 96 Simplify inside the parentheses: 2^(x+1)(4 - 1) = 96 Calculate: 2^(x+1)(3) = 96 Divide both sides by 3: 2^(x+1) = 32 Rewrite 32 as a power of 2: 2^(x+1) = 2^5 Set exponents equal: x + 1 = 5 Solve for x: x = 4 The answer is 4.
   Full step-by-step solution

   Step 1: Factor out 2^(x+1) from both terms: 2^(x+1)(2^2 - 1) = 96 Step 2: Simplify inside the parentheses: 2^(x+1)(4 - 1) = 96 Step 3: Calculate: 2^(x+1)(3) = 96 Step 4: Divide both sides by 3: 2^(x+1) = 32 Step 5: Rewrite 32 as a power of 2: 2^(x+1) = 2^5 Step 6: Set exponents equal: x + 1 = 5 Step 7: Solve for x: x = 4 The answer is 4.

2. A scientist is studying a population of algae in a pond. The algae population grows according to the function P(t) = 1200 × 2^(t/4), where P(t) is the population after t days. How many algae will be in the pond after 12 days? Answer: 9600 Solution: The population function is P(t) = 1200 × 2^(t/4) Substitute t = 12 into the function: P(12) = 1200 × 2^(12/4) Simplify the exponent: 12/4 = 3, so P(12) = 1200 × 2^3 Calculate 2^3 = 8 Multiply: 1200 × 8 = 9600 The algae population after 12 days is 9600.
   Full step-by-step solution

   Step 1: The population function is P(t) = 1200 × 2^(t/4) Step 2: Substitute t = 12 into the function: P(12) = 1200 × 2^(12/4) Step 3: Simplify the exponent: 12/4 = 3, so P(12) = 1200 × 2^3 Step 4: Calculate 2^3 = 8 Step 5: Multiply: 1200 × 8 = 9600 Step 6: The algae population after 12 days is 9600.

3. Liam is studying bacterial growth in his biology class. He starts with a colony of 500 bacteria that doubles every 3 hours. Using the exponential growth formula A = P(2)^(t/h), where A is the final amount, P is the initial population, t is time in hours, and h is the doubling time, how many bacteria will there be after 15 hours? Answer: 16000 Solution: Initial population P = 500 Doubling time h = 3 hours Time elapsed t = 15 hours Formula: A = P * (2)^(t/h) P = 500 t = 15 h = 3 t/h = 15 / 3 = 5 This means the bacteria double 5 times in 15 hours.
   Full step-by-step solution

   Let's solve this step-by-step. We are given: Initial population P = 500 Doubling time h = 3 hours Time elapsed t = 15 hours Formula: A = P * (2)^(t/h) --- **Step 1: Identify the values** P = 500 t = 15 h = 3 --- **Step 2: Calculate the exponent t/h** t/h = 15 / 3 = 5 This means the bacteria double 5 times in 15 hours. --- **Step 3: Substitute into the formula** A = 500 * (2)^5 --- **Step 4: Calculate 2^5** 2^5 = 2 * 2 * 2 * 2 * 2 = 32 --- **Step 5: Multiply** A = 500 * 32 500 * 30 = 15000 500 * 2 = 1000 15000 + 1000 = 16000 --- **Step 6: Final answer** After 15 hours, there will be 16000 bacteria.

4. Noah is comparing two functions to understand which type of growth eventually dominates. He considers the exponential function f(x) = 7^x and the quadratic function g(x) = x^10. For large values of x, which function grows faster? Show by evaluating both functions at x = 12 and x = 20, and explain how the comparison demonstrates that exponential growth eventually exceeds polynomial growth. Answer: f(x) = 7^x grows faster Solution: Evaluate f(x) = 7^x at x = 12 and x = 20. For x = 12: f(12) = 7^12. 7^2 = 49, 7^4 = 49^2 = 2401, 7^6 = 7^4 * 7^2 = 2401 * 49 = 117649, 7^12 = (7^6)^2 = 117649^2 = 13,841,287,201.
   Full step-by-step solution

   Step 1: Evaluate f(x) = 7^x at x = 12 and x = 20. For x = 12: f(12) = 7^12. 7^2 = 49, 7^4 = 49^2 = 2401, 7^6 = 7^4 * 7^2 = 2401 * 49 = 117649, 7^12 = (7^6)^2 = 117649^2 = 13,841,287,201. For x = 20: f(20) = 7^20 = (7^10)^2. 7^5 = 16807, 7^10 = 16807^2 = 282,475,249, so 7^20 = 282,475,249^2 = 79,792,266,297,612,001 (approximately 7.979 * 10^16). Step 2: Evaluate g(x) = x^10 at x = 12 and x = 20. For x = 12: g(12) = 12^10. 12^2 = 144, 12^5 = 12^2 * 12^2 * 12 = 144 * 144 * 12 = 248,832, 12^10 = (12^5)^2 = 248,832^2 = 61,917,364,224. For x = 20: g(20) = 20^10 = (2*10)^10 = 2^10 * 10^10 = 1024 * 10,000,000,000 = 10,240,000,000,000. Step 3: Compare at x = 12: f(12) = 13,841,287,201 and g(12) = 61,917,364,224. At x = 12, g(12) is larger than f(12). Step 4: Compare at x = 20: f(20) ≈ 79,792,266,297,612,001 and g(20) = 10,240,000,000,000. At x = 20, f(20) is about 7,792 times larger than g(20). Step 5: Conclusion: At x = 12, the quadratic function is still larger, but by x = 20, the exponential function has overtaken it dramatically. The ratio f(x)/g(x) changes from less than 1 at x = 12 to over 7,000 at x = 20. This shows that exponential growth (7^x) eventually far exceeds polynomial growth (x^10) because the exponential's repeated multiplication by a constant base outpaces the polynomial's fixed-degree growth for large x. The answer is f(x) = 7^x grows faster.

5. A biologist is studying a population of bacteria that grows exponentially. The initial population is 800 bacteria, and the population triples every 4 hours. Using the exponential growth model P(t) = P₀ × a^t, where P₀ is the initial population and a is the hourly growth factor, determine the population after 8 hours. Answer: 7200 Solution: Identify the given values: P₀ = 800 bacteria, tripling time = 4 hours, t = 8 hours Find the hourly growth factor a.
   Full step-by-step solution

   Step 1: Identify the given values: P₀ = 800 bacteria, tripling time = 4 hours, t = 8 hours Step 2: Find the hourly growth factor a. Since the population triples every 4 hours: a^4 = 3 Step 3: Calculate a: a = 3^(1/4) ≈ 1.31607 Step 4: Use the exponential growth formula: P(8) = 800 × (3^(1/4))^8 Step 5: Simplify the exponent: (3^(1/4))^8 = 3^(8/4) = 3^2 = 9 Step 6: Calculate the population: P(8) = 800 × 9 = 7200 Step 7: The population after 8 hours is 7200 bacteria.

6. A biologist is studying a population of bacteria that doubles every 3 hours. If the initial population is 500 bacteria, write an exponential function in the form P(t) = P₀ × a^t that models the population after t hours. What is the value of the growth factor a? Answer: 2^(1/3) Solution: - Initial population \( P_0 = 500 \) - Doubling time = 3 hours - Model: \( P(t) = P_0 \times a^t \), where \( t \) is in hours If the population doubles every 3 hours, then: P(3) = 2 \times P_0 P(3) = P_0 \times a^3 P_0 \times a^3 = 2 \times P_0 Cancel \( P_0 \) Since \( P_0 \neq 0 \), divide…
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Understand the problem** We know: - Initial population \( P_0 = 500 \) - Doubling time = 3 hours - Model: \( P(t) = P_0 \times a^t \), where \( t \) is in hours - Need to find \( a \) --- **Step 2: Use the doubling information** If the population doubles every 3 hours, then: \[ P(3) = 2 \times P_0 \] From the model: \[ P(3) = P_0 \times a^3 \] So: \[ P_0 \times a^3 = 2 \times P_0 \] --- **Step 3: Cancel \( P_0 \)** Since \( P_0 \neq 0 \), divide both sides by \( P_0 \): \[ a^3 = 2 \] --- **Step 4: Solve for \( a \)** \[ a = 2^{1/3} \] This is the cube root of 2. --- **Step 5: Write the final function** \[ P(t) = 500 \times (2^{1/3})^t \] Or equivalently: \[ P(t) = 500 \times 2^{t/3} \] --- **Step 6: Answer the question** The growth factor \( a \) is: \[ a = 2^{1/3} \] --- **Final answer:** \( 2^{1/3} \)

7. 2^(x+1) = 32, x = ? Answer: 4 Solution: Write 32 as a power of 2: 32 = 2^5 The equation becomes: 2^(x+1) = 2^5 Since the bases are equal, set the exponents equal: x + 1 = 5 Solve for x: x = 5 - 1 x = 4 The answer is 4.
   Full step-by-step solution

   Step 1: Write 32 as a power of 2: 32 = 2^5 Step 2: The equation becomes: 2^(x+1) = 2^5 Step 3: Since the bases are equal, set the exponents equal: x + 1 = 5 Step 4: Solve for x: x = 5 - 1 Step 5: x = 4 The answer is 4.

8. Compare f(x) = 8^x and g(x) = x^8. At what integer value of x > 7 does f(x) first exceed g(x)? Answer: 9 Solution: Test x = 8 f(8) = 8^8 = 16,777,216 g(8) = 8^8 = 16,777,216 At x = 8, f(x) = g(x) Test x = 9 f(9) = 8^9 = 134,217,728 g(9) = 9^8 = 43,046,721 At x = 9, f(9) > g(9) Since f(8) = g(8) and f(9) > g(9), the first integer value greater than 7 where the exponential exceeds the polynomial is x = 9.
   Full step-by-step solution

   Step 1: Test x = 8 f(8) = 8^8 = 16,777,216 g(8) = 8^8 = 16,777,216 At x = 8, f(x) = g(x) Step 2: Test x = 9 f(9) = 8^9 = 134,217,728 g(9) = 9^8 = 43,046,721 At x = 9, f(9) > g(9) Step 3: Since f(8) = g(8) and f(9) > g(9), the first integer value greater than 7 where the exponential exceeds the polynomial is x = 9. The answer is 9.