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.

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- Feb 26th 2010, 10:59 PMPeyton SawyerCollege Algebra help: exponential and logarithmic Functions
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. - Feb 26th 2010, 11:19 PMProve It
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 $\displaystyle n$. - Feb 26th 2010, 11:31 PMPeyton Sawyer
- Feb 26th 2010, 11:49 PMProve It
- Feb 26th 2010, 11:51 PMPeyton Sawyer
- Feb 27th 2010, 04:00 AMProve It
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 $\displaystyle t$.