The population of the world was 3 billlion in 1960 and it increased on average at annual of 18%...
In what year does the world population reach 5 billion ppl.
Year 0: $\displaystyle P = 3\cdot 10^9$
Year 1: $\displaystyle P = 1.18\cdot 3 \cdot 10^9$
Year 2: $\displaystyle P = 1.18\cdot 1.18 \cdot 3 \cdot 10^9$
Year 3: $\displaystyle P = 1.18 \cdot 1.18 \cdot 1.18 \cdot 3 \cdot 10^9$.
Can you see that the equation for Year $\displaystyle n$ would be
$\displaystyle P = 3\cdot 10^9 \cdot 1.18^n$.
Now let $\displaystyle P = 5 \cdot 10^9$ and solve for $\displaystyle n$.
For the first question:
$\displaystyle 5\cdot 10^9 = 3\cdot 10^9 \cdot 1.18^n$
$\displaystyle 1.18^n = \frac{5}{3}$
$\displaystyle \left(\frac{59}{50}\right)^n = \frac{5}{3}$
$\displaystyle \ln{\left(\frac{59}{50}\right)^n} = \ln{\frac{5}{3}}$
$\displaystyle n\ln{\frac{59}{50}} = \ln{\frac{5}{3}}$
$\displaystyle n = \frac{\ln{\frac{5}{3}}}{\ln{\frac{59}{50}}}$
$\displaystyle = \frac{\ln{5} - \ln{3}}{\ln{59} - \ln{50}}$.
For your second question:
If the interest is compounding continuously then
$\displaystyle A = Pe^{rt}$.
Here $\displaystyle A = 2P, r = \frac{7}{100}$.
Solve for $\displaystyle t$.