Exponential Growth

Question

I'm trying to wrap my head around the algebra used to get a solution. The question states:

In 2011, the Population of China and India were approximately 1.34 and 1.19 billion people, respectively. However due to central control the annual population growth rate of China was 0.4% while the population of India was growing by 1.37% each year. if these growth rates remain constant. when will the population of India exceed that of China?

• 2023

So the general formula would be $P = P(not) A^{kt}$ so I've tried $1.34 = 1.19e^{0.0137t}$ ---divide by 1.19 on both sides and take ln of both sides $\ln(1.34/1.19) = 0.0137t$. I, quiet cluelessly, divided by 0.0137 on both sides but that of course would give me an erroneous solution.

I generally understand exponential growth, or at least the idea behind how to calculate certain values, but this question in particular I haven't quiet understood. I would appreciate any help on how to go about correctly finding the correct value of t (2023). I'm sure my algebra skills are at fault

Answer 1

If the population grows at $1.37\%$ per year, after $n$ years it is multiplied by $1.0137^n$, so the population of India after $n$ years is $P(I)=1.19\cdot 1.0137^n$. Similarly the population of China is $P(C)=1.34\cdot 1.004^n$ You are asked to set these equal and solve for $n$. It appears your $1.43$ is a typo for $1.34$ and you have ignored the growth in China.

Answer 2

So the growth function for the population of China is $C(t)=1.34(1.004)^t$, and for India $I(t)=1.19(1.0137)^t$. So, we need to solve the inequality

$$\begin{align} 1.19(1.0137)^t & > 1.34(1.004)^t\\ \left(\frac{1.19}{1.34}\right)\left(\frac{1.0137}{1.004}\right)^t &>1\\ \left(\frac{1.0137}{1.004}\right)^t & > \left(\frac{1.34}{1.19}\right)\\ t\log\left(\frac{1.0137}{1.004}\right)& > \log\left(\frac{1.34}{1.19}\right)\\ t & > {\log(1.34/1.19)\over\log(1.0137/1.004)}\\ t& > 12.35 \end{align}$$

Answer 3

If you add $p\%$ each year, e.g. $p = 0.4 \% = 0.004$, then the initial population $P(0)$ grows to $$ P(k) = P(0) (1 + p)^k $$ after $k$ years. This is an exponential growth as well, not necessarily to base $e$. Solving for a suitable $k$ can be achieved by applying a logarithm: \begin{align} P_I(k) &> P_C(k) \iff \\ P_I(0) (1+p_I)^k &> P_C(0) (1+p_C)^k \Rightarrow \\ k \ln \frac{1+p_I}{1+p_C} & > \ln \frac{P_C(0)}{P_I(0)} \Rightarrow \\ k & > \frac{\ln \frac{P_C(0)}{P_I(0)}}{ \ln \frac{1+p_I}{1+p_C}} \end{align} This gives a relative time in terms of $k$ years, to verify against your result year you would need to know the year when the initial population values were taken.

Answer 4

$$1.34\cdot1.004^x=1.19\cdot1.0137^x\implies$$

$$\left(\frac{1.0137}{1.004}\right)^x=\frac{1.34}{1.19}\implies$$

$$x=\log_{\frac{1.0137}{1.004}}\frac{1.34}{1.19}\implies$$

$$x=\frac{\log\frac{1.34}{1.19}}{\log\frac{1.0137}{1.004}}\implies$$

$$x\approx12.347$$