Question

Geometric Question

Answer 1

1200(1-1.03^n)/(1-1.03) + 50,000 = 3600(1-1.03^n)/(1-1.03) + 10,000

Answer 2

FORMULAS INCOMING!

Answer 3

\[
\begin{align*}
\frac{1\,200(1-1.03^n)}{(1-1.03)}+50\,000&=\frac{3\,600(1-1.03^n)}{(1-1.03)}+10\,000
\end{align*}
\]
Is this formula correct? Just want to make sure not doing futile work.

Answer 4

@Leaper Is this correct?

Answer 5

Yes

Answer 6

Thomas I need to solve for n

Answer 7

I should of said so.

Answer 8

Thomas5267 my teacher said it is a two digit answer.

Answer 9

OK. My browser crashed. Now I have to type the whole thing again. Sorry.

Answer 10

Thats okay.

Answer 11

OK. Crashed again. Wait even more.

Answer 12

\[
\begin{align*}
\frac{1\,200(1-1.03^n)}{(1-1.03)}+50\,000&=\frac{3\,600(1-1.03^n)}{(1-1.03)}+10\,000 \\
\frac{1\,200(1-1.03^n)}{(1-1.03)}+40\,000&=\frac{3\,600(1-1.03^n)}{(1-1.03)} \\
\frac{(1-1.03^n)}{(1-1.03)}+\frac{40\,000}{1\,200}&=\frac{3(1-1.03^n)}{(1-1.03)} \\
\frac{(1-1.03^n)}{-0.03}+\frac{100}{3}&= \frac{3(1-1.03^n)}{-0.03} \\
1-1.03^n+\frac{-0.03\times100}{3}&=3(1-1.03^n) \\
1-1.03^n-1&=3-3(1.03^n) \\
-1.03^n&=3-3(1.03^n) \\
3(1.03^n)-1.03^n&=3 \\
2(1.03^n)&=3 \\
1.03^n&=1.5 \\
\log_{1.03}1.03^n&= \log_{1.03}1.5 \\
n&=\frac{\ln1.5}{\ln1.03} \\
n&=13.1712
\end{align*}
\]
I am not sure but at least Wolfram|Alpha says that I am correct.
http://www.wolframalpha.com/input/?i=Solve%5B%281200+%281+-+1.03%5En%29%29%2F%281+-+1.03%29+%2B+50000+%3D%3D+%28++++3600+%281+-+1.03%5En%29%29%2F%281+-+1.03%29+%2B+10000%2C+n%5D

Answer 13

@Leaper