GED Mathematical Reasoning TestAlgebraic Problem Solving with Expressions and EquationsHard
A scientist is studying a chemical reaction where the concentration of a reactant, C, in moles per liter, decreases over time 't' in minutes according to the function C(t) = 0.5e^(-0.1t). How long will it take for the concentration to reach 0.1 moles per liter?
- Aln(0.5) / -0.1 minutes
- Bln(5) / -0.1 minutes
- Cln(0.2) / -0.1 minutes
- Dln(0.1) / -0.1 minutes
Show answer & explanationAnswer & explanation
Correct answer: B. ln(5) / -0.1 minutes
Set C(t) = 0.1: 0.1 = 0.5e^(-0.1t). Divide by 0.5: 0.2 = e^(-0.1t). Take the natural logarithm of both sides: ln(0.2) = -0.1t. Solve for t: t = ln(0.2) / -0.1. This is equivalent to t = ln(1/5) / -0.1 = -ln(5) / -0.1 = ln(5) / 0.1. Since ln(0.2) is negative, dividing by -0.1 makes t positive. Alternatively, 0.2 = 1/5, so ln(1/5) = -ln(5). Thus, -ln(5) = -0.1t, which means t = ln(5)/0.1.
Why the other options are wrong
- A. This would be the result if the initial concentration was 0.1 and we were looking for when it reaches 0.5, or a miscalculation.
- C. This is the correct intermediate step, but the final simplification to a positive value is missing.
- D. This is incorrect; it seems to be taking ln of the target concentration directly without isolating the exponential term.
Solving Exponential Equations (Natural Log)
The process of finding the value of a variable in the exponent of an equation by using the natural logarithm (ln), especially when the base is 'e'.
- Isolate the exponential term first.
- Take the natural logarithm (ln) of both sides.
- Use the logarithm property ln(b^x) = x * ln(b) to bring the exponent down.
- Solve for the variable algebraically.
Memory trick: Exponential 'e': Isolate, then Natural Log it!