Exponential growth precalc population

Question

The population of City A increases by 8% every 10 years. The population of City B triples every 120 years. The two cities had equal populations of 10,000 residents each in the year 2000. In what year will city B have twice as many residents as city A?

Answer 1

Let $t$ the time in years with $t=0$ correspondig to year $2000$, $P_A(t)$ and $P_B(t)$ can be written by \begin{align} P_A(t)&=(1.08)^{t/10}P_0& \text{and} & & P_B(t)&=3^{t/120}P_0 \end{align} where $P_0=10000$, $P_B(t)=2P_A(t)\iff (1.08)^{t/10}=2\cdot3^{t/120}$, taking logs we have \begin{align} \frac{t}{120}\log3&=\log 2+\frac{t}{10}\log(1.08)\\ \left(\frac{1}{120}\log3-\frac{1}{10}\log(1.08)\right)t&=\log2\\ t&=\frac{120\log2}{\log3-12\log(1.08)}\\ &\approx 475. \end{align} Therefore, in year 2475 the population of City B will be twice that of the City A.

Answer 2

Here are the relevant formulas you need. $t$ is measured in years, and $t=0$ corresponds to the year 2000. $A(t)$ represents the population of city $A$, and $B(t)$ likewise for the other city.

$$A(t)=10000e^{k_1t}$$ $$B(t)=10000e^{k_2t}$$

Now, find the constants $k_1, k_2$ using the data you were given. Then, solve the equation $$B(t)=2A(t)$$ This is an equation entirely in $t$, after you've plugged in $k_1, k_2$. You may solve it by taking logs of both sides and rearranging. Find the value(s) of $t$ that make this hold. Then, translate to a year by adding 2000.