Exponential Functions and the Number e: Algebra 2 Math Test Answer Key – PDF Download

Algebra 2 Exponential Test Answer Key: questions 1–5

Question 1: Growth model (2 points)

1. Initial population 18,000; growth 4.5% per year.
2. \(P(8)=18000(1.045)^{8}\approx\) 25,598.

Scoring: 1 point for each part.

Question 2: Compound interest (2 points)

1. \(A=2400(1+\frac{0.032}{12})^{60}\).
2. The balance is $2,815.83.

Scoring: 1 point for the correct formula with n = 12 and nt = 60; 1 point for the correct balance.

Question 3: Continuous versus yearly (2 points)

1. \(A=1800e^{0.045\cdot 12}=1800e^{0.54}\approx\) $3,088.81.
2. \(A=1800(1.045)^{12}\approx\) $3,052.59, which is $36.23 less (using unrounded values).

Scoring: 1 point for each balance method with the matching difference in part (b).

Question 4: Model from a graph (3 points)

1. \(a=f(0)=5\) and \(b=\frac{15}{5}=3\). Check: \(\frac{45}{15}=3\).
2. \(f(x)=5\cdot 3^{x}\).
3. \(f(4)=5\cdot 81=\)405.

Scoring: 1 point for a and b, 1 point for the equation, 1 point for f(4) = 405.

Question 5: Solve for time (3 points)

1. \(1.04^{t}=1.75\), so \(t=\frac{\ln 1.75}{\ln 1.04}\).
2. \(t\approx\) 14.27.
3. After 14 years \(A(14)\approx\) $3,463.35 is below $3,500, and after 15 years \(A(15)\approx\) $3,601.89 is above. The answer is 15 years.

Scoring: 1 point for the exact logarithm form, 1 point for 14.27, 1 point for checking years 14 and 15.

Algebra 2 Exponential Test Answer Key: questions 6–8

Question 6: Doubling time (2 points)

Solve \(1.06^{t}=2\): \(t=\frac{\ln 2}{\ln 1.06}\approx\) 11.90 years.

Scoring: 1 point for the equation 1.06^t = 2 with a logarithm step; 1 point for the correct value.

Question 7: Half-life (3 points)

1. The mass halves from 240 to 120 in 5 days: 5 days.
2. \(A(t)=240(\frac{1}{2})^{t/5}\).
3. \(A(12)=240(\frac{1}{2})^{2.4}\approx\) 45.5 mg.

Scoring: 1 point for the half-life, 1 point for the model, 1 point for the value.

Question 8: Equivalent forms (3 points)

1. Since \(b=1.12\), the growth rate is 12% per year.
2. \(e^{k}=1.12\), so \(k=\ln 1.12\approx\) 0.1133, giving \(f(t)=500e^{0.1133t}\).
3. \(b_{\text{month}}=1.12^{1/12}\approx\) 1.00949, so the monthly rate is about 0.95%.

Scoring: 1 point for each part.

Scoring summary

How to use this algebra 2 exponential test answer key

Use this algebra 2 exponential test answer key to score the test fairly and to see which skills need more practice.

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Algebra 2 Exponential Functions and the Number e test answer key

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