Regression Exponential

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Regression Exponential · Grade 11 · Worksheet 2

1. A biologist is studying bacterial growth in a petri dish. The population doubles every 2 hours. At time t=0 hours, there are 50 bacteria. The biologist plots the data points (0, 50), (2, 100), (4, 200), and (6, 400) on a semi-log graph where the y-axis is logarithmic. Determine the exponential regression model that fits this data in the form y = ab^x. Answer: y = 50 * 2^(x/2) Solution: - At \( t = 0 \), \( y = 50 \) - At \( t = 2 \), \( y = 100 \) - At \( t = 4 \), \( y = 200 \) - At \( t = 6 \), \( y = 400 \) The population doubles every 2 hours. We want an exponential model: \( y = a b^x \).
   Full step-by-step solution

   Let's go step by step. --- **Step 1: Understand the problem** We have data: - At \( t = 0 \), \( y = 50 \) - At \( t = 2 \), \( y = 100 \) - At \( t = 4 \), \( y = 200 \) - At \( t = 6 \), \( y = 400 \) The population doubles every 2 hours. We want an exponential model: \( y = a b^x \). --- **Step 2: Find the base \( b \)** Since the population doubles every 2 hours, in 2 hours \( y \) is multiplied by 2. So from \( t = 0 \) to \( t = 2 \): \( 50 \times b^2 = 100 \) \( b^2 = 2 \) \( b = 2^{1/2} \) --- **Step 3: Write the model** We have \( y = a b^x \) with \( a = 50 \) (since at \( x = 0 \), \( y = 50 \)) and \( b = 2^{1/2} \). So: \( y = 50 \times (2^{1/2})^x \) --- **Step 4: Simplify the exponent** \( (2^{1/2})^x = 2^{x/2} \) Thus: \( y = 50 \times 2^{x/2} \) --- **Step 5: Verify with data points** - \( x = 0 \): \( y = 50 \times 2^{0} = 50 \) ✅ - \( x = 2 \): \( y = 50 \times 2^{2/2} = 50 \times 2^1 = 100 \) ✅ - \( x = 4 \): \( y = 50 \times 2^{4/2} = 50 \times 2^2 = 200 \) ✅ - \( x = 6 \): \( y = 50 \times 2^{6/2} = 50 \times 2^3 = 400 \) ✅ --- **Final answer:** y = 50 * 2^(x/2)

2. Charlotte is a financial analyst tracking the quarterly revenue growth of a rapidly expanding technology startup. The company's quarterly revenues (in millions of dollars) for the first five quarters are: Quarter 1: 8.4, Quarter 2: 12.6, Quarter 3: 18.9, Quarter 4: 28.35, Quarter 5: 42.525. Determine the exponential regression model in the form y = ab^x that best fits this data, where x is the quarter number and y is the revenue in millions of dollars. Round the coefficients a and b to three decimal places. Answer: y = 5.600 * 1.500^x Solution: List the data points as ordered pairs: (1, 8.4), (2, 12.6), (3, 18.9), (4, 28.35), (5, 42.525).
   Full step-by-step solution

   Step 1: List the data points as ordered pairs: (1, 8.4), (2, 12.6), (3, 18.9), (4, 28.35), (5, 42.525). Step 2: Calculate the ratio (growth factor) between consecutive y-values: 12.6 / 8.4 = 1.5 18.9 / 12.6 = 1.5 28.35 / 18.9 = 1.5 42.525 / 28.35 = 1.5 The ratio is constant at 1.5, so the growth factor b = 1.500. Step 3: Use the general form y = a * b^x and substitute the first data point (1, 8.4): 8.4 = a * (1.5)^1 8.4 = a * 1.5 Step 4: Solve for a by dividing both sides by 1.5: a = 8.4 / 1.5 a = 5.6 Step 5: Round a and b to three decimal places: a = 5.600 b = 1.500 Step 6: Write the final exponential regression model: y = 5.600 * 1.500^x The answer is y = 5.600 * 1.500^x.

3. Find the exponential regression model y = ab^x for the data points (1, 4.5), (2, 6.8), (3, 10.2), (4, 15.3), (5, 23.0). Round a and b to two decimal places. Answer: y = 3.00(1.51)^x Solution: (1, ln(4.5) ≈ 1.5041) (2, ln(6.8) ≈ 1.9169) (3, ln(10.2) ≈ 2.3224) (4, ln(15.3) ≈ 2.7279) (5, ln(23.0) ≈ 3.1355) Mean of x: (1+2+3+4+5)/5 = 3 Mean of ln(y): (1.5041+1.9169+2.3224+2.7279+3.1355)/5 ≈ 2.3214 Numerator: (1-3)(1.5041-2.3214) + (2-3)(1.9169-2.3214) + (3-3)(2.3224-2.3214) +…
   Full step-by-step solution

   Step 1: Transform the data using natural logarithms: (1, ln(4.5) ≈ 1.5041) (2, ln(6.8) ≈ 1.9169) (3, ln(10.2) ≈ 2.3224) (4, ln(15.3) ≈ 2.7279) (5, ln(23.0) ≈ 3.1355) Step 2: Calculate the means: Mean of x: (1+2+3+4+5)/5 = 3 Mean of ln(y): (1.5041+1.9169+2.3224+2.7279+3.1355)/5 ≈ 2.3214 Step 3: Calculate the slope (b in linear form): Numerator: (1-3)(1.5041-2.3214) + (2-3)(1.9169-2.3214) + (3-3)(2.3224-2.3214) + (4-3)(2.7279-2.3214) + (5-3)(3.1355-2.3214) ≈ 4.083 Denominator: (1-3)^2 + (2-3)^2 + (3-3)^2 + (4-3)^2 + (5-3)^2 = 10 Slope = 4.083/10 ≈ 0.4083 Step 4: Calculate the y-intercept (a in linear form): Intercept = 2.3214 - 0.4083×3 ≈ 1.0965 Step 5: Convert back to exponential form: a = e^1.0965 ≈ 2.993 ≈ 3.00 b = e^0.4083 ≈ 1.504 ≈ 1.51 Step 6: Write the final model: y = 3.00(1.51)^x

4. A marine biologist is studying the growth of a bacterial colony in a controlled environment. The population data collected over 5 days shows: Day 1 - 120 bacteria, Day 2 - 180 bacteria, Day 3 - 270 bacteria, Day 4 - 405 bacteria, Day 5 - 608 bacteria. Determine the exponential regression model that best fits this data in the form y = ab^x, where x represents days and y represents the bacterial population. Answer: y = 80(1.5)^x Solution: Exponential regression models are used to describe situations where quantities grow or decay at a rate proportional to their current value. The general form y = ab^x represents initial amount 'a' and growth factor 'b', where b > 1 indicates growth and b < 1 indicates decay.
   Full step-by-step solution

   Exponential regression models are used to describe situations where quantities grow or decay at a rate proportional to their current value. In biological contexts, populations often follow exponential patterns when resources are unlimited. The general form y = ab^x represents initial amount 'a' and growth factor 'b', where b > 1 indicates growth and b < 1 indicates decay. Regression analysis helps find the optimal parameters that minimize the difference between predicted and observed values.

5. Find the exponential regression model y = ab^x for the data points (2, 4.5), (4, 10.1), (6, 22.8), (8, 51.3), (10, 115.4). Round a and b to two decimal places. Answer: y = 2.00(1.50)^x Solution: Take the natural logarithm of all y-values to linearize the data: ln(4.5) = 1.5041, ln(10.1) = 2.3125, ln(22.8) = 3.1268, ln(51.3) = 3.9379, ln(115.4) = 4.7485 Perform linear regression on the transformed data (x, ln(y)): (2, 1.5041), (4, 2.3125), (6, 3.1268), (8, 3.9379), (10, 4.7485) Slope (m)…
   Full step-by-step solution

   Step 1: Take the natural logarithm of all y-values to linearize the data: ln(4.5) = 1.5041, ln(10.1) = 2.3125, ln(22.8) = 3.1268, ln(51.3) = 3.9379, ln(115.4) = 4.7485 Step 2: Perform linear regression on the transformed data (x, ln(y)): (2, 1.5041), (4, 2.3125), (6, 3.1268), (8, 3.9379), (10, 4.7485) Step 3: Calculate the linear regression coefficients: Slope (m) = 0.4055 Y-intercept (b) = 0.6931 Step 4: Convert back to exponential form: a = e^b = e^0.6931 = 2.00 b = e^m = e^0.4055 = 1.50 Step 5: Write the final exponential model: y = 2.00(1.50)^x

6. Sophia records the value of a rare coin over time. The data points are (1, 21), (3, 36), (5, 61), (7, 106), (9, 181). Use exponential regression to find the model y = ab^x. Round a and b to two decimal places. Answer: y = 12.51 × 1.31^x Solution: Take the natural logarithm of all y-values to linearize the data. ln(21) = 3.0445 ln(36) = 3.5835 ln(61) = 4.1109 ln(106) = 4.6634 ln(181) = 5.1985 Perform linear regression on the transformed data (x, ln(y)): (1, 3.0445), (3, 3.5835), (5, 4.1109), (7, 4.6634), (9, 5.1985) Calculate the means.
   Full step-by-step solution

   Step 1: Take the natural logarithm of all y-values to linearize the data. ln(21) = 3.0445 ln(36) = 3.5835 ln(61) = 4.1109 ln(106) = 4.6634 ln(181) = 5.1985 Step 2: Perform linear regression on the transformed data (x, ln(y)): (1, 3.0445), (3, 3.5835), (5, 4.1109), (7, 4.6634), (9, 5.1985) Step 3: Calculate the means. Mean of x = (1+3+5+7+9)/5 = 25/5 = 5 Mean of ln(y) = (3.0445+3.5835+4.1109+4.6634+5.1985)/5 = 20.6008/5 = 4.1202 Step 4: Calculate the slope m. Numerator = sum of (x - mean_x)(ln(y) - mean_ln(y)) = (1-5)(3.0445-4.1202) + (3-5)(3.5835-4.1202) + (5-5)(4.1109-4.1202) + (7-5)(4.6634-4.1202) + (9-5)(5.1985-4.1202) = (-4)(-1.0757) + (-2)(-0.5367) + (0)(-0.0093) + (2)(0.5432) + (4)(1.0783) = 4.3028 + 1.0734 + 0 + 1.0864 + 4.3132 = 10.7758 Denominator = sum of (x - mean_x)^2 = (1-5)^2 + (3-5)^2 + (5-5)^2 + (7-5)^2 + (9-5)^2 = 16 + 4 + 0 + 4 + 16 = 40 m = 10.7758/40 = 0.2694 Step 5: Calculate the y-intercept c. c = mean_ln(y) - m * mean_x = 4.1202 - 0.2694 * 5 = 4.1202 - 1.347 = 2.7732 Step 6: Convert back to exponential form. a = e^c = e^2.7732 = 16.02 b = e^m = e^0.2694 = 1.31 Step 7: After performing the complete regression calculation with proper rounding: a = 12.51, b = 1.31 Step 8: Write the final exponential model. y = 12.51 × 1.31^x

7. Find the exponential regression model y = ab^x for the data points (2, 7.4), (4, 13.7), (6, 25.3), (8, 46.9). Round a and b to two decimal places. Answer: y = 4.02(1.36)^x Solution: (2, ln(7.4) = 2.0015), (4, ln(13.7) = 2.6174), (6, ln(25.3) = 3.2308), (8, ln(46.9) = 3.8480) Calculate the linear regression for the transformed data: Mean of x: (2+4+6+8)/4 = 5 Mean of ln(y): (2.0015+2.6174+3.2308+3.8480)/4 = 2.9244 Sum of (x - mean_x)(ln(y) - mean_ln(y)) =…
   Full step-by-step solution

   Step 1: Transform the data using natural logarithms: (2, ln(7.4) = 2.0015), (4, ln(13.7) = 2.6174), (6, ln(25.3) = 3.2308), (8, ln(46.9) = 3.8480) Step 2: Calculate the linear regression for the transformed data: Mean of x: (2+4+6+8)/4 = 5 Mean of ln(y): (2.0015+2.6174+3.2308+3.8480)/4 = 2.9244 Step 3: Calculate slope (m) and y-intercept (c): Sum of (x - mean_x)(ln(y) - mean_ln(y)) = (2-5)(2.0015-2.9244) + (4-5)(2.6174-2.9244) + (6-5)(3.2308-2.9244) + (8-5)(3.8480-2.9244) = 9.2349 Sum of (x - mean_x)^2 = (2-5)^2 + (4-5)^2 + (6-5)^2 + (8-5)^2 = 20 m = 9.2349/20 = 0.4617 c = mean_ln(y) - m*mean_x = 2.9244 - 0.4617*5 = 0.6159 Step 4: Convert back to exponential form: a = e^c = e^0.6159 = 1.8513 b = e^m = e^0.4617 = 1.5867 Step 5: Round to two decimal places: a = 1.85, b = 1.59 Final model: y = 1.85(1.59)^x