# Population of a city doubles in $50$ years. In how many years will it triple?

The population of a city doubles in $$50$$ years. In how many years will it triple, under the assumption that the rate of increase is proportional to the number of inhabitants?

This was a pretty trivial question. But, the thing is that when I tried solving this as:

We have, the population of a city becoming double in $$50$$ years. This means, that the population increases by $$100$$ percent in $$50$$ years. This hints at the fact that the population was increasing at the rate of $$2$$ percent per year. So, we consider the initial population as $$P$$.

Let the number of years in which it becomes triple is $$y$$ years. Then, we have the following equation holding true,

$$\left(P+\frac{2Py}{100}\right)=3P\implies y=100\text{years}.$$

So, I thought, the population will become thrice in $$100$$ years.

But here, comes the problem. The answer given suggests the population becomes thrice in $$50\log_23$$ years. I don't understand the how is it so? Have I done some error/or forgotten to take something into consideration?

• If $P$ is the population, in 50 years it will be $2P$, and in 100 years it will be $4P$, since it doubled again. So tripling is somewhere between 50 and 100 years, don't you agree? May 12 at 7:31
• Exponential growth. Off the top of my head log base 2 of 3 times 50. May 12 at 7:36
• Just making a quick guess at your error based upon a scan, 100% increase in the first 50 years takes it from 100% to 200%. 100% increase in the 2nd 50 years takes it from 200% to 400%. Because it's 100% of the population at 50 years, not 100% of the population at the outset. May 12 at 7:38
• Also, welcome to Maths Stack Exchange! And thank-you for formatting your question well - many don't get it right from the outset. Check the edit I made to see how to make taller brackets. May 12 at 7:41
• Your equation does not hold true. Rather, the equation you want to consider (provided your previous assumptions hold true) is $P(1+0.02)^y=3P.$ But the idea here is that the correct model for the population is of the form $Pe^{ry},$ where $r$ is a constant that may be found from the fact that $Pe^{50r}=2P,$ and you may verify that indeed you get $50\log_2 3.$ May 12 at 7:44

Well, in this context, the city's population grows exponentially, not linearly (which means as time goes on, the rate of growth increases/decreases in a curve and doesn't stay at a single value). You'd be right if it does grow linearly – but it doesn't!

Now, we can think of the population as a function of time, something like $$n\cdot2^t$$ where $$t$$ is the time and $$n$$ is the initial population.

Let's say that $$t$$ is measured in a half-century, or 50 years (so that $$t=2$$ represents 100 years, etc). For each 50 years that pass by ($$t=1$$), the population will increase from $$n$$ to $$n\cdot2^1$$, hence doubling the population. So, this way, we can find what the population is for a given $$t$$.

Consider this: if we know the population but want to find the $$t$$ when the city reaches that population, we're basically asking “what $$t$$ fulfills that $$n\cdot2^t=something$$?” For your scenario $$n\cdot2^t=3n$$, we can simply change that to $$2^t=3$$.

Now, to directly access $$t$$'s value here, we can use the $$\log$$ function (as you probably know).

So, $$2^t=3$$ can be written as $$t=\log_2(3)$$. Since $$t$$ represents 50 years, the number of years it'll take to triple the population is simply $$50t$$. 🙂

• Yup! Much clearer now.
– Stef
May 12 at 9:37
• @TheMather-orratherAMather This is indeed a brilliant answer! I have just two questions, though. $(1)$ What motivated you to take $f(t)=n2^t$. You could have chosen any $x$ instead of $2$. My understanding is that, since it's given the population doubles in $50$ years, so, you thought choosing $2$ was the most appropriate ,is it ? May 12 at 12:03
• @TheMather-orratherAMather $(2)$ Why did you explicitly choose the function of $t$ of the $n2^t$ (I am again asking the motivation behind this). The function could also have been of the form say $n/e^t,logn^t,$ etc ? My understanding for this motivation is:From the link you suggested en.m.wikipedia.org/wiki/Exponential_growth , the definition of exponential growth seems to be, that a quantity say, $x_0$, grows exponentially, then it's value at discrete time intervals $t=1,2,3,...$ is $x_t=x_0(1+r)^t,$ where $r$ is a real constant and it seems this is the definition of exponential May 12 at 12:11
• growth which you used and asserted that the function of $t$ will be $n(2)^t$ , where you used $r=1$ and $x_0$ as $n$. But then, you choice for $r$ to be $1$ was due to my understanding for my question $(1)$, correct ? (i.e by question $(1)$ I mean the question $(1)$ describe here math.stackexchange.com/questions/4697617/…, in my previous comment) May 12 at 12:13
• @ThomasFinley nice analysis! Yes, I chose $2$ because you said the population doubles every 50 years. You're right that we can use any $x$. Based on the information we have, though, $2$ is the most "convenient" one to use (for instance, I'd use $3$ if you said something like "the population triples every 80 years" instead). Also, yes that we can use $e$ and some others instead, but I decided to keep it simple (rather than using, say, $n\cdot e^{\ln(2)t}$). I suppose this problem isn't that complicated, and plus, I'm definitely not the best at doing more advanced stuff like those 🙂 May 12 at 12:49

The statement "under the assumption that the rate of increase is proportional to the number of inhabitants" means that the function number of inhabitants $$P(t)$$, where $$t$$ is the time (time unit = 1 year), solves the linear differential equation $$\frac{dP}{dt}(t)=c P(t)$$ where $$c$$ is the constant of proportionality. The general solution of such equation is $$P(t)=P(0)\exp(ct)$$ where $$P(0)$$ is the initial population. If $$P(0)>0$$ and $$c>0$$ then the population $$P(t)$$ grows exponentially.

Now "the population becoming double in 50 years" implies that $$P(t+50)=2P(t).$$ From this equation we can find the constant $$c$$: assuming $$P(0)\not=0$$, we get $$P(0)\exp(c(t+50))=2P(0)\exp(ct)\implies \exp(50c)=2\implies c=\frac{\ln(2)}{50}.$$ Finally we find $$T$$="number of years to triple" by solving $$P(t+T)=3P(t)$$ which yields $$P(0)\exp(c(t+T))=3P(0)\exp(ct)\implies \exp(cT)=3\\\implies T=\frac{\ln(3)}{c}=50\ln_2(3)\approx 79.25 years.$$

More generally, for $$n\geq 1$$, the equation $$P(t+T_n)=nP(t)$$ implies $$T_n=50\ln_2(n)$$, which means that at ANY time $$t\geq 0$$, given a population of $$P(t)$$, after $$T_n$$ years the population will be $$n$$ times $$P(t)$$.

This hints at the fact that the population was increasing at the rate of 2 percent per year.

This assumes a crude approximation at each timestep, and that percentages are additive, not multiplicative. (In fact, your logic will generally always result in this answer, regardless of how you divide up the time intervals.)

Suppose, in a strange world, that I am given two discounts on a $$\100$$ bill: a $$50\%$$ discount, and a $$25\%$$ discount. Assume these are applied in order to the bill, and affect the total price of the bill post discounts.

Your logic seems to suggest a $$75\%$$ discount in the end, and a final bill of $$\25$$. However, you would have $$\100 \cdot 0.5 \cdot 0.75 = \37.50$$ instead.

If you have a population of $$100$$ people that grows at $$2\%$$ per year, then you have $$100 \cdot 1.02 = 102$$ people after one year; $$102 \cdot 1.02 = 100 \cdot 1.02^2$$ after two years; $$100 \cdot 1.02^3$$ after three, and so on. However, if you do this for $$100$$ years, you get $$100 \cdot 1.02^{100} \approx 724$$ people. On the other hand, a population that grows by $$2 \times 100\%=200\%$$ after $$100$$ years only has $$100 \cdot 2 = 200$$ people.

This is how exponential growth essentially works: growth based on a multiple of the previous population. Loosely, $$P(t+\Delta t) \approx P(t) + \Delta t P'(t)$$ where $$P$$ is the population, and $$t$$ a time variable. (Assume $$\Delta t$$ is small.) Phrased differently, $$\frac{dP}{dt} = k P$$ for some constant $$k$$: the rate at which the number grows is proportional to the value itself. Note in particular we have to use continuously varying quantities in this process.

In your model, you start with a population $$P$$, and poof $$P/50$$ people into existence the very moment $$y$$ years is hit, when in reality those people are having children the entire time -- and in theory, those children are themselves also having children, and their children are too. (Think of the Fibonacci bunnies.) You need to account for this continual growth more appropriately to get an accurate answer.

Another analogy you may find useful is the difference between interest compounded $$n$$ times per year, versus continuously-compounded interest.

Solve the differential equation given for the appropriate $$k$$ to figure out a closed form for $$P(t)$$, and then you can handle matters accordingly.

Simply put, you're assuming a linear rate of growth, while the problem asks you to assume an exponential rate of growth. Say for example the city starts with 10,000 people. Since it grows to 20,000 people in 50 years, the average rate of growth per year over that timespan (the slope of the secant line between the two data points) is 200 people per year. But it is a mistake to assume that the city grows at a constant and unchanging rate of growth of 200 people per year no matter how much time passes. Instead, the question asks you to model the population exponentially so that the rate of growth at any point in time is proportional to the population at that time. The model you have proposed is $$P(t) = P + 0.02Pt$$, which is an equation describing the population of a city that has a constant influx of the same amount of people every year equal to 2% of the original population. Instead, the correct model should be $$P(t) = P \times 2^{(t/50)}$$, which is an equation describing the population of a city that doubles in population every 50 years. Plugging in the appropriate values into this model yields the correct answer.

This means, that the population increases by 100 percent in 50 years. This hints at the fact that the population was increasing at the rate of 2 percent per year.

Wrong. If the population increased by 2 percent each year, then it would be $$x$$ at the start, then $$1.02x$$ after year one, $$1.02^2$$ after year $$2$$ and so on, and it would be $$1.02^{50}x$$ after $$50$$ years. Since $$1.02^{50} \approx 2.69$$, this population would more than double in $$50$$ years, which contradicts your assumptions.

A simple way is to envisage a compounding annual growth rate $$R$$, say, for the population,

then $$R^{50} =2\tag 1$$

the time $$T$$ we need is $$R^T = 3\tag2$$

and dividing $$(2)$$ by $$(1)$$ and taking logs of both and simplifying,

$$T = 50\cdot\large\frac{log 3}{log 2}\,, \approx 79.25$$ years