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Exponential Growth

Grade 9 · Algebra · Worksheet 3

  1. A research scientist named Maya is studying the spread of a new virus in a city with a population of 50,000 people. The virus spreads such that the number of infected people triples every 5 days. If there are initially 20 infected people, write an exponential function in the form I(t) = I₀ × b^t that models the number of infected people after t days, where I₀ is the initial number and b is the daily growth factor. Then, determine the daily growth factor b. Answer: ______________
  2. Tane is analyzing two functions on a coordinate plane: the exponential function f(x) = 2^x and the quadratic function g(x) = x^2. He notices that for small values of x, the quadratic appears to be above the exponential. However, he wants to determine the exact integer x-values greater than 8 where the exponential function's output first becomes greater than the quadratic's output, and then continues to stay greater for all larger x. Find the smallest integer x greater than 8 such that 2^x > x^2, and confirm that this inequality holds for the next two consecutive integer x-values as well. Answer: ______________
  3. Mason is comparing two functions on a coordinate plane. The first function is f(x) = 9^x, and the second function is g(x) = x^9. He plots both graphs for x ≥ 0. For what smallest integer value of x does the graph of f(x) first exceed the graph of g(x)? Answer: ______________
  4. On a coordinate plane, Aroha graphs two functions: f(x) = 9^x and g(x) = x^9. For what smallest integer value of x greater than 1 does the exponential function f(x) first exceed the polynomial function g(x)? Answer: ______________
  5. A scientist is studying a population of algae in a pond. The algae population grows exponentially 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: ______________
  6. On a coordinate plane, Tane graphs two functions: the exponential function f(x) = 10^x and the polynomial function g(x) = x^10. For what smallest integer value of x greater than 1 does the exponential function f(x) first exceed the polynomial function g(x)? Answer: ______________
  7. 2^5 - 3^2 = ? Answer: ______________
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Answer Key & Explanations

Exponential Growth · Grade 9 · Worksheet 3

  1. A research scientist named Maya is studying the spread of a new virus in a city with a population of 50,000 people. The virus spreads such that the number of infected people triples every 5 days. If there are initially 20 infected people, write an exponential function in the form I(t) = I₀ × b^t that models the number of infected people after t days, where I₀ is the initial number and b is the daily growth factor. Then, determine the daily growth factor b. Answer: 3^(1/5) Solution: Identify the initial number of infected people, I₀ = 20. The number of infected people triples every 5 days. This means the growth factor over 5 days is 3.
    Full step-by-step solution

    Step 1: Identify the initial number of infected people, I₀ = 20. Step 2: The number of infected people triples every 5 days. This means the growth factor over 5 days is 3. Step 3: To find the daily growth factor b, we need the number such that b^5 = 3. Step 4: Therefore, b = 3^(1/5). The daily growth factor is 3^(1/5).

  2. Tane is analyzing two functions on a coordinate plane: the exponential function f(x) = 2^x and the quadratic function g(x) = x^2. He notices that for small values of x, the quadratic appears to be above the exponential. However, he wants to determine the exact integer x-values greater than 8 where the exponential function's output first becomes greater than the quadratic's output, and then continues to stay greater for all larger x. Find the smallest integer x greater than 8 such that 2^x > x^2, and confirm that this inequality holds for the next two consecutive integer x-values as well. Answer: 9, 10, 11 Solution: Test x = 8 (just to see the trend). f(8) = 2^8 = 256, g(8) = 8^2 = 64. Here 256 > 64, so the exponential is already greater at x = 8.
    Full step-by-step solution

    Step 1: Test x = 8 (just to see the trend). f(8) = 2^8 = 256, g(8) = 8^2 = 64. Here 256 > 64, so the exponential is already greater at x = 8. But the problem asks for x greater than 8, so we start at x = 9. Step 2: Test x = 9. f(9) = 2^9 = 512, g(9) = 9^2 = 81. 512 > 81, so the inequality holds at x = 9. Step 3: Test x = 10. f(10) = 2^10 = 1024, g(10) = 10^2 = 100. 1024 > 100, so the inequality holds at x = 10. Step 4: Test x = 11. f(11) = 2^11 = 2048, g(11) = 11^2 = 121. 2048 > 121, so the inequality holds at x = 11. Step 5: For x > 11, the exponential grows by a factor of 2 each step while the quadratic grows by a diminishing relative amount, so the inequality will continue to hold for all larger x. Thus, the smallest integer x greater than 8 is 9, and the inequality holds for x = 9, 10, and 11. The answer is 9, 10, 11.

  3. Mason is comparing two functions on a coordinate plane. The first function is f(x) = 9^x, and the second function is g(x) = x^9. He plots both graphs for x ≥ 0. For what smallest integer value of x does the graph of f(x) first exceed the graph of g(x)? Answer: 10 Solution: We need to find the smallest integer x >= 0 such that 9^x > x^9. x = 0: 9^0 = 1, 0^9 = 0. 1 > 0, so f(x) exceeds g(x) at x=0.
    Full step-by-step solution

    We need to find the smallest integer x >= 0 such that 9^x > x^9. Let's test integer values of x: x = 0: 9^0 = 1, 0^9 = 0. 1 > 0, so f(x) exceeds g(x) at x=0. But we are looking for the first time it exceeds after a possible initial crossover. Let's continue to see the pattern. x = 1: 9^1 = 9, 1^9 = 1. 9 > 1. x = 2: 9^2 = 81, 2^9 = 512. 81 < 512. Here g(x) is larger. x = 3: 9^3 = 729, 3^9 = 19683. 729 < 19683. x = 4: 9^4 = 6561, 4^9 = 262144. 6561 < 262144. x = 5: 9^5 = 59049, 5^9 = 1953125. 59049 < 1953125. x = 6: 9^6 = 531441, 6^9 = 10077696. 531441 < 10077696. x = 7: 9^7 = 4782969, 7^9 = 40353607. 4782969 < 40353607. x = 8: 9^8 = 43046721, 8^9 = 134217728. 43046721 < 134217728. x = 9: 9^9 = 387420489, 9^9 = 387420489. They are equal. x = 10: 9^10 = 3486784401, 10^9 = 1000000000. 3486784401 > 1000000000. At x = 10, f(x) = 9^10 = 3,486,784,401 and g(x) = 10^9 = 1,000,000,000. Since 3,486,784,401 > 1,000,000,000, the exponential function f(x) = 9^x exceeds the polynomial function g(x) = x^9 at x = 10. Therefore, the smallest integer x where f(x) first exceeds g(x) is 10. The answer is 10.

  4. On a coordinate plane, Aroha graphs two functions: f(x) = 9^x and g(x) = x^9. For what smallest integer value of x greater than 1 does the exponential function f(x) first exceed the polynomial function g(x)? Answer: 10 Solution: Evaluate f(x) = 9^x and g(x) = x^9 for x = 2: f(2) = 9^2 = 81, g(2) = 2^9 = 512. Here, f(2) < g(2). For x = 3: f(3) = 9^3 = 729, g(3) = 3^9 = 19683.
    Full step-by-step solution

    Step 1: Evaluate f(x) = 9^x and g(x) = x^9 for x = 2: f(2) = 9^2 = 81, g(2) = 2^9 = 512. Here, f(2) < g(2). Step 2: For x = 3: f(3) = 9^3 = 729, g(3) = 3^9 = 19683. f(3) < g(3). Step 3: For x = 4: f(4) = 9^4 = 6561, g(4) = 4^9 = 262144. f(4) < g(4). Step 4: For x = 5: f(5) = 9^5 = 59049, g(5) = 5^9 = 1953125. f(5) < g(5). Step 5: For x = 6: f(6) = 9^6 = 531441, g(6) = 6^9 = 10077696. f(6) < g(6). Step 6: For x = 7: f(7) = 9^7 = 4782969, g(7) = 7^9 = 40353607. f(7) < g(7). Step 7: For x = 8: f(8) = 9^8 = 43046721, g(8) = 8^9 = 134217728. f(8) < g(8). Step 8: For x = 9: f(9) = 9^9 = 387420489, g(9) = 9^9 = 387420489. f(9) = g(9). Step 9: For x = 10: f(10) = 9^10 = 3486784401, g(10) = 10^9 = 1000000000. f(10) > g(10). Thus, the smallest integer x greater than 1 where f(x) exceeds g(x) is x = 10. The answer is 10.

  5. A scientist is studying a population of algae in a pond. The algae population grows exponentially 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.

  6. On a coordinate plane, Tane graphs two functions: the exponential function f(x) = 10^x and the polynomial function g(x) = x^10. For what smallest integer value of x greater than 1 does the exponential function f(x) first exceed the polynomial function g(x)? Answer: 11 Solution: Evaluate f(x) = 10^x and g(x) = x^10 for x = 2: f(2) = 10^2 = 100, g(2) = 2^10 = 1024. Here, f(2) < g(2). For x = 3: f(3) = 10^3 = 1000, g(3) = 3^10 = 59049.
    Full step-by-step solution

    Step 1: Evaluate f(x) = 10^x and g(x) = x^10 for x = 2: f(2) = 10^2 = 100, g(2) = 2^10 = 1024. Here, f(2) < g(2). Step 2: For x = 3: f(3) = 10^3 = 1000, g(3) = 3^10 = 59049. f(3) < g(3). Step 3: For x = 4: f(4) = 10^4 = 10000, g(4) = 4^10 = 1048576. f(4) < g(4). Step 4: For x = 5: f(5) = 10^5 = 100000, g(5) = 5^10 = 9765625. f(5) < g(5). Step 5: For x = 6: f(6) = 10^6 = 1000000, g(6) = 6^10 = 60466176. f(6) < g(6). Step 6: For x = 7: f(7) = 10^7 = 10000000, g(7) = 7^10 = 282475249. f(7) < g(7). Step 7: For x = 8: f(8) = 10^8 = 100000000, g(8) = 8^10 = 1073741824. f(8) < g(8). Step 8: For x = 9: f(9) = 10^9 = 1000000000, g(9) = 9^10 = 3486784401. f(9) < g(9). Step 9: For x = 10: f(10) = 10^10 = 10000000000, g(10) = 10^10 = 10000000000. f(10) = g(10). Step 10: For x = 11: f(11) = 10^11 = 100000000000, g(11) = 11^10 = 25937424601. f(11) > g(11). Thus, the smallest integer x greater than 1 where f(x) exceeds g(x) is x = 11. The answer is 11.

  7. 2^5 - 3^2 = ? Answer: 23 Solution: Calculate 2^5 2^5 means 2 multiplied by itself 5 times: 2 × 2 = 4 4 × 2 = 8 8 × 2 = 16 16 × 2 = 32 So, 2^5 = 32. Calculate 3^2 3^2 means 3 multiplied by itself 2 times: 3 × 3 = 9 So, 3^2 = 9. We have 2^5 - 3^2 = 32 - 9.
    Full step-by-step solution

    Let's solve the problem step by step. Step 1: Calculate 2^5 2^5 means 2 multiplied by itself 5 times: 2 × 2 = 4 4 × 2 = 8 8 × 2 = 16 16 × 2 = 32 So, 2^5 = 32. Step 2: Calculate 3^2 3^2 means 3 multiplied by itself 2 times: 3 × 3 = 9 So, 3^2 = 9. Step 3: Subtract the results We have 2^5 - 3^2 = 32 - 9. 32 - 9 = 23. Final Answer: 23