Solve the problem.
-A certain country's population $\mathrm{P}(\mathrm{t})$, in millions, tyears after 2010 can be approximated by $\mathrm{P}(\mathrm{t})=3.495(1.018)^{\mathrm{t}}$.
Find the doubling time. Round your answer to the nearest tenth.
A) $38.9 \mathrm{yr}$
B) $61.6 \mathrm{yr}$
C) $77.7 \mathrm{yr}$
D) $70.1 \mathrm{yr}$

Question

Quizplus · Accepted Answer

To find the doubling time, set $P(t) = 2 \times 3.495$ and solve for $t$. Using the formula $P(t) = 3.495(1.018)^t$, we get $2 = (1.018)^t$. Taking the natural logarithm of both sides and solving for $t$ gives a value close to 38.9 years.

Solve the Problem P(t)\mathrm{P}(\mathrm{t})P(t) , in Millions, Tyears After 2010 Can Be Approximated By Millions

Solve the Problem $\mathrm{P}(\mathrm{t})$ , in Millions, Tyears After 2010 Can Be Approximated By Millions