Exponential Growth 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
 A: 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
$$\begin{align}
1.19(1.0137)^t & > 1.34(1.004)^t\\
\left(\frac{1.19}{1.34}\right)\left(\frac{1.0137}{1.004}\right)^t &>1\\
\left(\frac{1.0137}{1.004}\right)^t & > \left(\frac{1.34}{1.19}\right)\\
t\log\left(\frac{1.0137}{1.004}\right)& > \log\left(\frac{1.34}{1.19}\right)\\
t & > {\log(1.34/1.19)\over\log(1.0137/1.004)}\\
t& > 12.35
\end{align}$$
A: 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.
A: $$1.34\cdot1.004^x=1.19\cdot1.0137^x\implies$$
$$\left(\frac{1.0137}{1.004}\right)^x=\frac{1.34}{1.19}\implies$$
$$x=\log_{\frac{1.0137}{1.004}}\frac{1.34}{1.19}\implies$$
$$x=\frac{\log\frac{1.34}{1.19}}{\log\frac{1.0137}{1.004}}\implies$$
$$x\approx12.347$$
A: 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.
