# College Algebra help: exponential and logarithmic Functions

• Feb 26th 2010, 10:59 PM
Peyton Sawyer
College 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 PM
Prove It
Quote:

Originally Posted by Peyton Sawyer
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: $P = 3\cdot 10^9$

Year 1: $P = 1.18\cdot 3 \cdot 10^9$

Year 2: $P = 1.18\cdot 1.18 \cdot 3 \cdot 10^9$

Year 3: $P = 1.18 \cdot 1.18 \cdot 1.18 \cdot 3 \cdot 10^9$.

Can you see that the equation for Year $n$ would be

$P = 3\cdot 10^9 \cdot 1.18^n$.

Now let $P = 5 \cdot 10^9$ and solve for $n$.
• Feb 26th 2010, 11:31 PM
Peyton Sawyer
Quote:

Originally Posted by Prove It
Year 0: $P = 3\cdot 10^9$

Year 1: $P = 1.18\cdot 3 \cdot 10^9$

Year 2: $P = 1.18\cdot 1.18 \cdot 3 \cdot 10^9$

Year 3: $P = 1.18 \cdot 1.18 \cdot 1.18 \cdot 3 \cdot 10^9$.

Can you see that the equation for Year $n$ would be

$P = 3\cdot 10^9 \cdot 1.18^n$.

Now let $P = 5 \cdot 10^9$ and solve for $n$.

Determine how long it takes for an investment to double its value if the interest rate is 7% compounded continuously. Show equation set up.
• Feb 26th 2010, 11:49 PM
Prove It
Quote:

Originally Posted by Peyton Sawyer
Determine how long it takes for an investment to double its value if the interest rate is 7% compounded continuously. Show equation set up.

Also, did you manage to finish the last question?
• Feb 26th 2010, 11:51 PM
Peyton Sawyer
Quote:

Originally Posted by Prove It

Also, did you manage to finish the last question?

its part B to the first question.
And im still working on finishing it up. its sort of difficult
• Feb 27th 2010, 04:00 AM
Prove It
For the first question:

$5\cdot 10^9 = 3\cdot 10^9 \cdot 1.18^n$

$1.18^n = \frac{5}{3}$

$\left(\frac{59}{50}\right)^n = \frac{5}{3}$

$\ln{\left(\frac{59}{50}\right)^n} = \ln{\frac{5}{3}}$

$n\ln{\frac{59}{50}} = \ln{\frac{5}{3}}$

$n = \frac{\ln{\frac{5}{3}}}{\ln{\frac{59}{50}}}$

$= \frac{\ln{5} - \ln{3}}{\ln{59} - \ln{50}}$.

$A = Pe^{rt}$.
Here $A = 2P, r = \frac{7}{100}$.
Solve for $t$.