Exponential Growth Math Example 4

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Example 4

hard
Two populations grow exponentially: P1(t)=100⋅3tP_1(t) = 100 \cdot 3^t and P2(t)=5000⋅(1.2)tP_2(t) = 5000 \cdot (1.2)^t. When will P1P_1 overtake P2P_2?

Solution

  1. 1
    Set equal: 100⋅3t=5000⋅(1.2)t100 \cdot 3^t = 5000 \cdot (1.2)^t. So 3t(1.2)t=50\frac{3^t}{(1.2)^t} = 50, giving (31.2)t=50\left(\frac{3}{1.2}\right)^t = 50, i.e., (2.5)t=50(2.5)^t = 50.
  2. 2
    t=ln⁡50ln⁡2.5=3.9120.9163≈4.27t = \frac{\ln 50}{\ln 2.5} = \frac{3.912}{0.9163} \approx 4.27.

Answer

t≈4.27t \approx 4.27
When comparing two exponential functions, the one with the larger base always eventually dominates, regardless of initial values. The crossover point is found by setting them equal and solving. Here P1P_1 starts much smaller but its growth rate (200%200\% per unit) far exceeds P2P_2's (20%20\% per unit).

About Exponential Growth

Exponential growth occurs when a quantity increases by a constant multiplicative factor over equal intervals.

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