Exponential Growth

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Answer Key & Explanations

Exponential Growth · Grade 9 · Worksheet 2

1. Noah is studying two different growth models for a science fair project. The exponential function f(x) = 6^x models the spread of a population of bacteria, while the quadratic function g(x) = x^6 models the growth of a plant's height in centimeters. For large values of x, which function grows faster and why? Show by evaluating both functions at x = 6 and x = 11, and explain how the comparison demonstrates that exponential growth eventually exceeds polynomial growth. Answer: f(x) = 6^x grows faster Solution: Evaluate f(x) = 6^x at x = 6 and x = 11. For x = 6: f(6) = 6^6. Calculate 6^2 = 36, 6^3 = 36 * 6 = 216, 6^4 = 216 * 6 = 1296, 6^5 = 1296 * 6 = 7776, 6^6 = 7776 * 6 = 46656.
   Full step-by-step solution

   Step 1: Evaluate f(x) = 6^x at x = 6 and x = 11. For x = 6: f(6) = 6^6. Calculate 6^2 = 36, 6^3 = 36 * 6 = 216, 6^4 = 216 * 6 = 1296, 6^5 = 1296 * 6 = 7776, 6^6 = 7776 * 6 = 46656. For x = 11: f(11) = 6^11 = 6^6 * 6^5 = 46656 * 7776 = 362,797,056. Step 2: Evaluate g(x) = x^6 at x = 6 and x = 11. For x = 6: g(6) = 6^6 = 46656. For x = 11: g(11) = 11^6. Calculate 11^2 = 121, 11^3 = 121 * 11 = 1331, 11^4 = 1331 * 11 = 14641, 11^5 = 14641 * 11 = 161051, 11^6 = 161051 * 11 = 1,771,561. Step 3: Compare at x = 6: f(6) = 46656 and g(6) = 46656. They are equal at x = 6. Step 4: Compare at x = 11: f(11) = 362,797,056 and g(11) = 1,771,561. f(11) is about 204.8 times larger than g(11). Step 5: Conclusion: As x increases from 6 to 11, the exponential function f(x) = 6^x grows from being equal to the quadratic to being over 200 times larger. This shows that exponential growth eventually far exceeds polynomial growth because the exponential function's base (6) causes repeated multiplication that outpaces the polynomial's fixed power. The answer is f(x) = 6^x grows faster.

2. On a coordinate plane, Liam graphs two functions: y = 3^x and y = x^3. For x = 9, which function has the greater value? Then, consider the graphs for very large x values (like x = 15). Which function will eventually dominate and grow faster? Answer: y = 3^x Solution: Evaluate both functions at x = 9. For y = 3^x, we calculate 3^9. 3^1=3, 3^2=9, 3^3=27, 3^4=81, 3^5=243, 3^6=729, 3^7=2187, 3^8=6561, 3^9=19683.
   Full step-by-step solution

   Step 1: Evaluate both functions at x = 9. For y = 3^x, we calculate 3^9. 3^1=3, 3^2=9, 3^3=27, 3^4=81, 3^5=243, 3^6=729, 3^7=2187, 3^8=6561, 3^9=19683. For y = x^3, we calculate 9^3 = 9 * 9 * 9 = 81 * 9 = 729. At x = 9, 3^x (19683) is much larger than x^3 (729). Step 2: Evaluate both functions at x = 15. For y = 3^x, we calculate 3^15. 3^10 = 59049, 3^11 = 177147, 3^12 = 531441, 3^13 = 1594323, 3^14 = 4782969, 3^15 = 14348907. For y = x^3, we calculate 15^3 = 15 * 15 * 15 = 225 * 15 = 3375. Step 3: Compare the growth. At x = 9, 3^9 (19683) is about 27 times larger than 9^3 (729). At x = 15, 3^15 (14,348,907) is about 4,251 times larger than 15^3 (3375). The ratio of the exponential function's value to the cubic function's value is increasing dramatically as x grows. Step 4: Conclusion. Exponential functions (like 3^x) have a constant multiplicative growth factor, which eventually outpaces any polynomial function (like x^3), regardless of the polynomial's degree. For large values of x, y = 3^x will always be greater than y = x^3 and will grow at a much faster rate. The answer is y = 3^x.

3. Charlotte is comparing two investment options for a school project. Option A grows linearly: its value after x years is given by f(x) = 12x + 100. Option B grows exponentially: its value after x years is given by g(x) = 2^x. She wants to determine after how many complete years the exponential investment will first exceed the linear investment. At what year does this happen? Answer: Year 10 Solution: Set up the inequality to find when exponential exceeds linear: 2^x > 12x + 100. x = 1: f(1) = 12(1) + 100 = 112, g(1) = 2^1 = 2. 2 > 112 is false.
   Full step-by-step solution

   Step 1: Set up the inequality to find when exponential exceeds linear: 2^x > 12x + 100. Step 2: Test small values: x = 1: f(1) = 12(1) + 100 = 112, g(1) = 2^1 = 2. 2 > 112 is false. x = 5: f(5) = 12(5) + 100 = 160, g(5) = 2^5 = 32. 32 > 160 is false. x = 8: f(8) = 12(8) + 100 = 196, g(8) = 2^8 = 256. 256 > 196 is true. Step 3: Since it becomes true at x = 8, check x = 7: f(7) = 12(7) + 100 = 184, g(7) = 2^7 = 128. 128 > 184 is false. Step 4: Therefore, the exponential investment first exceeds the linear investment at year 8. Final answer: Year 8.

4. 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.

5. Emma is investigating the long-term behavior of two functions for her math project. She is comparing the exponential function f(x) = 3^x with the linear function g(x) = 15x + 45. Emma wants to determine the smallest positive integer value of x for which the exponential function f(x) first exceeds the linear function g(x). Find this value of x. Answer: 5 Solution: Set up the inequality to find when exponential exceeds linear: 3^x > 15x + 45. Test integer values of x starting from x = 1: x = 1: f(1) = 3^1 = 3, g(1) = 15(1) + 45 = 60. 3 > 60 is false.
   Full step-by-step solution

   Step 1: Set up the inequality to find when exponential exceeds linear: 3^x > 15x + 45. Step 2: Test integer values of x starting from x = 1: x = 1: f(1) = 3^1 = 3, g(1) = 15(1) + 45 = 60. 3 > 60 is false. x = 2: f(2) = 3^2 = 9, g(2) = 15(2) + 45 = 75. 9 > 75 is false. x = 3: f(3) = 3^3 = 27, g(3) = 15(3) + 45 = 90. 27 > 90 is false. x = 4: f(4) = 3^4 = 81, g(4) = 15(4) + 45 = 105. 81 > 105 is false. x = 5: f(5) = 3^5 = 243, g(5) = 15(5) + 45 = 120. 243 > 120 is true. Step 3: Since the inequality becomes true at x = 5, and we have checked all smaller positive integers, the smallest integer value of x where the exponential function first exceeds the linear function is x = 5. The answer is 5.

6. Ava is analyzing two different investment plans for her mathematics project. Plan A grows linearly, represented by the function L(x) = 20x + 150, where x is the number of years and L(x) is the total value in dollars. Plan B grows exponentially, represented by the function E(x) = 3^x. Ava wants to find the smallest whole number of years x for which the exponential plan's value first exceeds the linear plan's value. Determine this year. Answer: Year 6 Solution: Set up the inequality to find when exponential exceeds linear: 3^x > 20x + 150. Test x = 1: L(1) = 20(1) + 150 = 170, E(1) = 3^1 = 3. 3 > 170 is false.
   Full step-by-step solution

   Step 1: Set up the inequality to find when exponential exceeds linear: 3^x > 20x + 150. Step 2: Test x = 1: L(1) = 20(1) + 150 = 170, E(1) = 3^1 = 3. 3 > 170 is false. Step 3: Test x = 2: L(2) = 20(2) + 150 = 190, E(2) = 3^2 = 9. 9 > 190 is false. Step 4: Test x = 3: L(3) = 20(3) + 150 = 210, E(3) = 3^3 = 27. 27 > 210 is false. Step 5: Test x = 4: L(4) = 20(4) + 150 = 230, E(4) = 3^4 = 81. 81 > 230 is false. Step 6: Test x = 5: L(5) = 20(5) + 150 = 250, E(5) = 3^5 = 243. 243 > 250 is false. Step 7: Test x = 6: L(6) = 20(6) + 150 = 270, E(6) = 3^6 = 729. 729 > 270 is true. Step 8: Since at x = 5 the exponential was still smaller, and at x = 6 it becomes larger, the smallest whole number year is 6. Final answer: Year 6.

7. Compare f(x)=3^x and g(x)=x^3. For which x > 7 does f(x) > g(x)? Answer: 8 Solution: Evaluate f(7) = 3^7 = 2187 Evaluate g(7) = 7^3 = 343 At x=7, f(7)=2187 > g(7)=343, but we need to check if this is the first time f(x) exceeds g(x) Evaluate f(6) = 3^6 = 729 Evaluate g(6) = 6^3 = 216 At x=6, f(6)=729 > g(6)=216 Evaluate f(5) = 3^5 = 243 Evaluate g(5) = 5^3 = 125 At x=5, f(5)=243…
   Full step-by-step solution

   Step 1: Evaluate f(7) = 3^7 = 2187 Step 2: Evaluate g(7) = 7^3 = 343 Step 3: At x=7, f(7)=2187 > g(7)=343, but we need to check if this is the first time f(x) exceeds g(x) Step 4: Evaluate f(6) = 3^6 = 729 Step 5: Evaluate g(6) = 6^3 = 216 Step 6: At x=6, f(6)=729 > g(6)=216 Step 7: Evaluate f(5) = 3^5 = 243 Step 8: Evaluate g(5) = 5^3 = 125 Step 9: At x=5, f(5)=243 > g(5)=125 Step 10: Evaluate f(4) = 3^4 = 81 Step 11: Evaluate g(4) = 4^3 = 64 Step 12: At x=4, f(4)=81 > g(4)=64 Step 13: Evaluate f(3) = 3^3 = 27 Step 14: Evaluate g(3) = 3^3 = 27 Step 15: At x=3, f(3)=27 = g(3)=27 Step 16: Evaluate f(2) = 3^2 = 9 Step 17: Evaluate g(2) = 2^3 = 8 Step 18: At x=2, f(2)=9 > g(2)=8 Step 19: Evaluate f(1) = 3^1 = 3 Step 20: Evaluate g(1) = 1^3 = 1 Step 21: At x=1, f(1)=3 > g(1)=1 Step 22: The smallest x > 7 where f(x) > g(x) is actually x=8, but we need to check if f(x) > g(x) for ALL x > 7 Step 23: Since exponential functions eventually grow faster than polynomial functions, once f(x) exceeds g(x), it stays greater for all larger x values Step 24: Therefore, for all x > 7, f(x) > g(x) The answer is 8.