Grade 8 · Algebra & functions

Evaluating exponential functions practice

Evaluating exponential functions is a grade 8 math skill aligned to Common Core standard 8.EE.A.3: use numbers expressed in the form of a single digit times an integer power of 10 to estimate very large or very small quantities. Below are 8 practice questions with answers and step-by-step explanations, drawn from the 10 evaluating exponential functions problems our math games drill.

CCSS 8.EE.A.310 questions in the bank
Sample questions

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Question 1medium

Let f(t)=8e3t+50f(t) = 8e^{3t} + 50. Which of the following approximations, written as a×10na \times 10^n with one significant digit, is closest to the value of f(2)f(2)?

Substitute t=2t = 2: f(2)=8e6+50f(2) = 8e^{6} + 50. Since e6403.4e^{6} \approx 403.4, f(2)8(403.4)+50=3,2773×103f(2) \approx 8(403.4) + 50 = 3{,}277 \approx 3 \times 10^{3}.

Question 2medium

Let f(t)=4e3t+200f(t) = 4e^{3t} + 200. Which of the following approximations, written as a×10na \times 10^n with one significant digit, is closest to the value of f(3)f(3)?

Substitute t=3t = 3: f(3)=4e9+200f(3) = 4e^{9} + 200. Since e98,103e^{9} \approx 8{,}103, f(3)4(8,103)+200=32,6123×104f(3) \approx 4(8{,}103) + 200 = 32{,}612 \approx 3 \times 10^{4}.

Question 3medium

Let f(t)=4e2t+50f(t) = 4e^{2t} + 50. Which of the following approximations, written as a×10na \times 10^n with one significant digit, is closest to the value of f(4)f(4)?

Substitute t=4t = 4: f(4)=4e8+50f(4) = 4e^{8} + 50. Since e82,981e^{8} \approx 2{,}981, f(4)4(2,981)+50=11,9741×104f(4) \approx 4(2{,}981) + 50 = 11{,}974 \approx 1 \times 10^{4}.

Question 4hard

Let f(t)=3e2t+10f(t) = 3e^{2t} + 10. Which of the following approximations, written as a×10na \times 10^n with one significant digit, is closest to the value of f(5)f(5)?

Substitute t=5t = 5: f(5)=3e10+10f(5) = 3e^{10} + 10. Since e1022,026e^{10} \approx 22{,}026, f(5)3(22,026)+10=66,0887×104f(5) \approx 3(22{,}026) + 10 = 66{,}088 \approx 7 \times 10^{4}.

Question 5hard

Let f(t)=6e2tf(t) = 6e^{2t}. Which of the following approximations, written as a×10na \times 10^n with one significant digit, is closest to the value of f(5)f(5)?

Substitute t=5t = 5: f(5)=6e10f(5) = 6e^{10}. Since e1022,026e^{10} \approx 22{,}026, f(5)6(22,026)=132,1561×105f(5) \approx 6(22{,}026) = 132{,}156 \approx 1 \times 10^{5}.

Question 6hard

Let f(t)=5e3t+200f(t) = 5e^{3t} + 200. Which of the following approximations, written as a×10na \times 10^n with one significant digit, is closest to the value of f(4)f(4)?

Substitute t=4t = 4: f(4)=5e12+200f(4) = 5e^{12} + 200. Since e12162,755e^{12} \approx 162{,}755, f(4)5(162,755)+200=813,9758×105f(4) \approx 5(162{,}755) + 200 = 813{,}975 \approx 8 \times 10^{5}.

Question 7hard

Let f(t)=2e3t+1,000f(t) = 2e^{3t} + 1{,}000. Which of the following approximations, written as a×10na \times 10^n with one significant digit, is closest to the value of f(4)f(4)?

Substitute t=4t = 4: f(4)=2e12+1,000f(4) = 2e^{12} + 1{,}000. Since e12162,755e^{12} \approx 162{,}755, f(4)2(162,755)+1,000=326,5103×105f(4) \approx 2(162{,}755) + 1{,}000 = 326{,}510 \approx 3 \times 10^{5}.

Question 8hard

Let f(t)=e3tf(t) = e^{3t}. Which of the following approximations, written as a×10na \times 10^n with one significant digit, is closest to the value of f(5)f(5)?

Substitute t=5t = 5: f(5)=e15f(5) = e^{15}. Since e153,269,017e^{15} \approx 3{,}269{,}017, f(5)3×106f(5) \approx 3 \times 10^{6}.

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