Question

exponential and logarithmic functions
choosing an exponential model and using it to make a prediction
figure 1

y
2500
2000
1500
1000
500
0
5 10 15 20
x

y = 1250(0.91)^x
figure 2

y
2500
2000
1500
1000
500
0
5 10 15 20
x

y = 2476(0.92)^x
figure 3

y
2500
2000
1500
1000
500
0
5 10 15 20
x

y = -75x + 1600
(a) which curve fits the data best?
○ figure 1
○ figure 2
○ figure 3
(b) use the equation of the best fitting curve from part (a) to predict the area that the forest covers after it is inhabited for 13 years. round your answer to the nearest hundredth.
□ square kilometers

Explanation:

Part (a)

Step1: Analyze Figure 1

The curve in Figure 1 is \( y = 1250(0.91)^x \). The data points seem to have a larger spread from the curve, especially at lower \( x \)-values.

Step2: Analyze Figure 2

The curve in Figure 2 is \( y = 2476(0.92)^x \). Visually, the data points lie closer to this curve, following its exponential decay trend more closely than the other figures.

Step3: Analyze Figure 3

The curve in Figure 3 is a linear function \( y=-75x + 1600 \), but the data points show a curved (exponential) pattern, so the linear curve does not fit well.

Step1: Identify the function

From part (a), the best - fitting curve is \( y = 2476(0.92)^x \), where \( x \) represents the number of years and \( y \) represents the area of the forest in square kilometers.

Step2: Substitute \( x = 13 \) into the function

We need to calculate \( y=2476\times(0.92)^{13} \). First, calculate \( (0.92)^{13} \). Using a calculator, \( 0.92^{13}\approx0.3758 \). Then, multiply by 2476: \( y = 2476\times0.3758\approx2476\times0.3758 = 2476\times(0.3 + 0.07+0.005 + 0.0008)=2476\times0.3+2476\times0.07 + 2476\times0.005+2476\times0.0008=742.8+173.32 + 12.38+1.9808 = 930.4808\approx930.48 \) (using a more accurate calculator for \( 0.92^{13} \): \( 0.92^{13}=\mathrm{e}^{13\ln(0.92)}\approx\mathrm{e}^{13\times(- 0.0833)}\approx\mathrm{e}^{-1.0829}\approx0.338 \), then \( 2476\times0.338 = 2476\times0.3+2476\times0.03 + 2476\times0.008=742.8+74.28+19.808 = 836.888\)? Wait, no, let's use a calculator for precise calculation. \( 0.92^{13}\):

\( 0.92^1 = 0.92\)

\( 0.92^2=0.92\times0.92 = 0.8464\)

\( 0.92^3=0.92\times0.8464 = 0.778688\)

\( 0.92^4=0.92\times0.778688 = 0.71639296\)

\( 0.92^5=0.92\times0.71639296 = 0.6590815232\)

\( 0.92^6=0.92\times0.6590815232 = 0.6063549913\)

\( 0.92^7=0.92\times0.6063549913 = 0.5578465920\)

\( 0.92^8=0.92\times0.5578465920 = 0.5132188646\)

\( 0.92^9=0.92\times0.5132188646 = 0.4721613554\)

\( 0.92^{10}=0.92\times0.4721613554 = 0.4343884470\)

\( 0.92^{11}=0.92\times0.4343884470 = 0.3996373712\)

\( 0.92^{12}=0.92\times0.3996373712 = 0.3676663815\)

\( 0.92^{13}=0.92\times0.3676663815 = 0.3382530700\)

Then \( y = 2476\times0.3382530700\approx2476\times0.3383\)

\( 2476\times0.3383=(2400 + 76)\times0.3383=2400\times0.3383+76\times0.3383=811.92+25.7108 = 837.6308\approx837.63\)

Answer:

Figure 2

Part (b)