Question

we recommend using the desmos graphing calculator. answers have 5% tolerance. assume data fits a power equation: ( y = ax^b ).

x	y
50	78.5
70	92.9
107	115
125	yellow box

Explanation:

Step1: Take natural log of power equation

Given \( y = Ax^b \), take the natural logarithm of both sides: \( \ln(y)=\ln(A)+b\ln(x) \). Let \( Y = \ln(y) \), \( X=\ln(x) \), \( B = \ln(A) \), so the equation becomes \( Y = B + bX \), a linear equation.

Step2: Calculate X and Y for given data

For \( (x,y)=(50,78.5) \): \( X_1=\ln(50)\approx3.9120 \), \( Y_1=\ln(78.5)\approx4.3635 \)
For \( (x,y)=(70,92.9) \): \( X_2=\ln(70)\approx4.2485 \), \( Y_2=\ln(92.9)\approx4.5323 \)
For \( (x,y)=(107,115) \): \( X_3=\ln(107)\approx4.6729 \), \( Y_3=\ln(115)\approx4.7449 \)

Step3: Find slope \( b \) and intercept \( B \)

Using linear regression (or two - point formula, here we can use least - squares or average rate). Let's use two - point between first and third:
Slope \( b=\frac{Y_3 - Y_1}{X_3 - X_1}=\frac{4.7449 - 4.3635}{4.6729 - 3.9120}=\frac{0.3814}{0.7609}\approx0.5 \) (approximate, more accurately with all points: using linear regression, the slope \( b\approx0.5 \), intercept \( B=\ln(A)\approx Y - bX \). Taking first point: \( B = 4.3635-0.5\times3.9120 = 4.3635 - 1.956=2.4075 \), so \( A = e^{2.4075}\approx11.2 \))

Step4: Predict y for x = 125

First, find \( X_4=\ln(125)\approx4.8283 \)
Then \( Y_4=B + bX_4=2.4075+0.5\times4.8283=2.4075 + 2.41415 = 4.82165 \)
Then \( y = e^{Y_4}=e^{4.82165}\approx124.5 \) (with 5% tolerance, let's check with power equation \( y = Ax^b \), if \( A\approx11.2 \), \( b = 0.5 \), then \( y=11.2\times\sqrt{125}=11.2\times11.1803\approx125.22 \). Another way: using the trend, from x = 50 (y = 78.5), x = 70 (y = 92.9), x = 107 (y = 115), the power law \( y = Ax^{0.5}\) (since \( b\approx0.5 \)) fits. So \( y = A\sqrt{x} \), from x = 50, \( 78.5 = A\sqrt{50}\Rightarrow A=\frac{78.5}{\sqrt{50}}\approx\frac{78.5}{7.071}\approx11.1 \). Then for x = 125, \( y = 11.1\times\sqrt{125}=11.1\times11.1803\approx124.1 \). With 5% tolerance, the value should be around 124 - 126.

Answer:

\( \approx124.5 \) (or in the range 118.3 - 130.7 due to 5% tolerance)