Purpose: Self-check the same-base method, growth/decay models, and compound interest setups.
Score: 0 / 12
Question 1
Solve: \( 2^x = 32 \).
Solution:
\( 32 = 2^5 \), so \( x = 5 \).
Question 2
Solve: \( 3^{x+1} = 81 \).
Solution:
\( 81 = 3^4 \), so \( x + 1 = 4 \), \( x = 3 \).
Question 3
Solve: \( 4^x = 8 \).
Solution:
\( (2^2)^x = 2^3 \) so \( 2x = 3 \), \( x = 1.5 \).
Question 4
Solve: \( 5^{2x-1} = 125 \).
Solution:
\( 125 = 5^3 \), so \( 2x - 1 = 3 \), \( x = 2 \).
Question 5
Solve: \( 9^{x+1} = 27^{x-1} \).
Solution:
\( 3^{2(x+1)} = 3^{3(x-1)} \), so \( 2x + 2 = 3x - 3 \), \( x = 5 \).
Question 6
A bacterial culture doubles every 30 minutes. Starting at 200 bacteria, the population after 2 hours is:
Solution:
2 hours = 4 doubling periods. \( 200 \cdot 2^4 = 200 \cdot 16 = 3200 \).
Question 7
Iodine-131 has a half-life of 8 days. If a sample is 100 mg, the mass after 32 days is:
Solution:
32 days = 4 half-lives. \( 100 \cdot (\tfrac{1}{2})^4 = 100/16 = 6.25 \) mg.
Question 8
A population grows at 5% per year, starting at 50{,}000. The population after 4 years is approximately:
Solution:
\( 50000(1.05)^4 \approx 50000 \cdot 1.2155 \approx 60{,}775 \).
Question 9
\$2000 invested at 4% per year compounded annually for 6 years gives (round to nearest dollar):
Solution:
\( A = 2000(1.04)^6 \approx 2000 \cdot 1.2653 \approx \$2530 \).
Question 10
For 6% per year compounded quarterly for 3 years, the periodic rate \( i \) and number of periods \( n \) are:
Solution:
Quarterly: \( i = 0.06/4 = 0.015 \), \( n = 3 \cdot 4 = 12 \).
Question 11
An investment doubles in approximately how many years at 8% compounded annually? (Use the rule of 72.)
Solution:
Rule of 72: \( 72/8 = 9 \) years.
Question 12
A drug has half-life 6 hours. Starting at 200 mg, the amount after 18 hours is:
Solution:
18 hours = 3 half-lives. \( 200 \cdot (\tfrac{1}{2})^3 = 200/8 = 25 \) mg.