Digital SATMath: AlgebraHard
A financial analyst is comparing two investment options. Option A's value, V_A(t), is modeled by V_A(t) = 1000(1.05)^t, and Option B's value, V_B(t), is modeled by V_B(t) = 1500(1.03)^t, where t is the number of years. Approximately how many years will it take for the value of Option A to exceed the value of Option B?
- A30 years
- B20 years
- C35 years
- D25 years
Show answer & explanationAnswer & explanation
Correct answer: D. 25 years
We need to find t such that V_A(t) > V_B(t), so 1000(1.05)^t > 1500(1.03)^t. Divide both sides by 1000(1.03)^t: (1.05)^t / (1.03)^t > 1500/1000. This simplifies to (1.05/1.03)^t > 1.5, or (1.0194)^t > 1.5. Take the natural logarithm of both sides: t * ln(1.0194) > ln(1.5). So, t > ln(1.5) / ln(1.0194) ≈ 0.405 / 0.0192 ≈ 21.09. Therefore, it will take approximately 25 years for Option A to exceed Option B.
Why the other options are wrong
- A. This value is too high and would mean Option A has significantly surpassed Option B.
- B. This value is too low; at 20 years, Option A is still slightly less than Option B.
- C. This value is significantly too high, indicating a large margin of error in calculation.
Comparing Exponential Functions
To find when one exponential function exceeds another, set up an inequality and solve for the variable, often requiring the use of logarithms.
- Set the inequality: f(t) > g(t) or f(t) < g(t).
- Isolate the exponential terms on one side.
- Use logarithms to solve for the exponent variable.
Memory trick: Compare Exponentials: Set inequality, simplify bases, then log both sides!