Digital SATMath: Advanced MathMedium
A scientist is modeling the population growth of a certain bacteria culture. The population P(t) after t hours is given by P(t) = P_0 * e^(kt), where P_0 is the initial population and k is the growth rate constant. If the initial population is 1,000 bacteria and the population doubles every 3 hours, what is the value of the growth rate constant k?
- Aln(3) / 2
- B3 / ln(2)
- C2 / 3
- Dln(2) / 3
Show answer & explanationAnswer & explanation
Correct answer: D. ln(2) / 3
If the population doubles every 3 hours, then P(3) = 2 * P_0. Substitute this into the formula: 2P_0 = P_0 * e^(k*3). Divide by P_0: 2 = e^(3k). Take the natural logarithm of both sides: ln(2) = 3k. Solve for k: k = ln(2)/3.
Why the other options are wrong
- A. This option uses incorrect values for the doubling time and the logarithm.
- B. This option inverts the relationship of ln(2) and 3.
- C. This results from an incorrect transformation of the exponential equation, possibly by ignoring the logarithm.
Exponential Growth Rate Constant
In exponential growth models (P(t) = P_0 * e^(kt)), the growth rate constant 'k' determines how quickly the quantity increases over time. It can be found using known doubling time or half-life.
- For doubling time (T_d): 2 = e^(k * T_d) which implies k = ln(2) / T_d.
- For half-life (T_h): 0.5 = e^(k * T_h) which implies k = ln(0.5) / T_h = -ln(2) / T_h (for decay).
- The constant 'k' is positive for growth and negative for decay.
Memory trick: Doubling time to 'k', use ln(2) divided by three!