# Do I have the formula right? How do I solve for n?

• Mar 23rd 2009, 04:20 PM
kcwillia377
Do I have the formula right? How do I solve for n?
A population of 8 million increases at a rate of .1% every month. In how many months will it reach 20 million?

Is the formula:

20000000=8000000*(1.01)^(n-1)
• Mar 23rd 2009, 04:24 PM
e^(i*pi)
Quote:

Originally Posted by kcwillia377
A population of 8 million increases at a rate of .1% every month. In how many months will it reach 20 million?

Is the formula:

20000000=8000000*(1.01)^(n-1)

I think it's ^n instead of ^(n-1)

$\displaystyle 20 \times 10^6 = 8 \times 10^6 (1.01)^n$
• Mar 23rd 2009, 04:24 PM
guardofthecolor4ever
yeah that's the correct formula...i'm pretty sure...but i could be wrong
• Mar 23rd 2009, 04:32 PM
kcwillia377
How do I solve for n?
• Mar 23rd 2009, 04:54 PM
e^(i*pi)
Quote:

Originally Posted by e^(i*pi)
I think it's ^n instead of ^(n-1)

$\displaystyle 20 \times 10^6 = 8 \times 10^6 (1.01)^n$

Quote:

Originally Posted by kcwillia377
How do I solve for n?

$\displaystyle 1.01^n = \frac{5}{2}$

Take logs:

$\displaystyle n\ln(1.01) = ln{\frac{5}{2}}$

$\displaystyle n = \frac{ln(2.5)}{ln(1.01)} = 92.1$