Solve the problem.
-A certain country's population $\mathrm { P } ( \mathrm { t } )$, in millions, t years after 1980 can be approximated by $\mathrm { P } ( \mathrm { t } ) = 2.495 ( 1.018 ) ^ { \mathrm { t } } \text {. }$ Find the doubling time. Round your answer to the nearest tenth.
A) $61.6 \mathrm { yr }$
B) $38.9 \mathrm { yr }$
C) $77.7 \mathrm { yr }$
D) $51.2 \mathrm { yr }$

Question

Quizplus · Accepted Answer

We need to find the time it takes for the population to double, which means we need to solve the equation:
$$2P(t) = P(0) \cdot (1.018)^t$$
where $P(0) = 2.495$. Simplifying this equation, we get:
$$2 \cdot 2.495 = 2.495 \cdot (1.018)^t$$
$$ 1.995984 = (1.018)^t$$
To solve for $t$, we take the natural logarithm of both sides:
$$\ln{(1.995984)} = t \ln{(1.018)}$$
$$t = \frac{\ln{(1.995984)}}{\ln{(1.018)}} \approx 38.9$$
Therefore, the doubling time is approximately 38.9 years, which corresponds to choice B.

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