tboss Posts: 4, Reputation: 1 New Member #1 Mar 19, 2007, 04:20 PM
Solve in terms of n
Hi All, I have this equation.
$f=c((1+r)^n + \frac{(1+r)^{n+1}-(1+r)}{r})$

I'm trying to solve in terms of n (i.e. making n the subject of the formula. Does anyone have any idea how this can be done? Thanks a lot.
 tboss Posts: 4, Reputation: 1 New Member #2 Mar 20, 2007, 01:45 AM
Hi guys, someone please throw me a lifeline!!
 tboss Posts: 4, Reputation: 1 New Member #3 Mar 20, 2007, 02:01 PM
I've heard that MathCad can be used to solve this... I don't have MathCad. Can someone try it out?
 tboss Posts: 4, Reputation: 1 New Member #4 Mar 20, 2007, 02:02 PM
 galactus Posts: 2,271, Reputation: 282 Ultra Member #5 Mar 20, 2007, 02:05 PM
I ran this through my trusty ol' 92 and got:

$n=\frac{ln\left(\frac{(c+f)r+c}{c(2r+1)}\right)}{l n(r+1)}$

$\frac{(c+f)r+c}{c(2r+1)}>0$
 asterisk_man Posts: 476, Reputation: 32 Full Member #6 Mar 21, 2007, 11:36 AM
OK. Simple. (once I've shown you ;) )

$f=c\left({\left(1+r\right)}^n + \frac{{\left(1+r\right)}^{n+1}-\left(1+r\right)}{r}\right) \\

\text{divide by c} \\
\frac f c={\left(1+r\right)}^n + \frac{{\left(1+r\right)}^{n+1}-\left(1+r\right)}{r} \\

\text{multiply by r} \\
\frac {fr} c=r{\left(1+r\right)}^n + {\left(1+r\right)}^{n+1}-\left(1+r\right) \\

\text{expand the middle term} \\
\frac {fr} c=r{\left(1+r\right)}^n + {\left(1+r\right)}^n\left(1+r\right)-\left(1+r\right) \\

\text{collect the } \left(1+r\right)^n \text{terms}\\
\frac {fr} c=\left(r+\left(1+r\right)\right){\left(1+r\right) }^n -\left(1+r\right) \\

\text{simplify } r+(1+r) \text{and add} (1+r) \\
\frac {fr} c + \left(1+r\right) = \left( 2r+1\right){\left(1+r\right)}^n \\

\text{divide by } 2r+1 \\
\frac {\frac {fr} c + \left(1+r\right)} {\left( 2r+1\right)} = {\left(1+r\right)}^n \\

\text{simplify the left by multiplying by } \frac c c \\
\frac {fr + c + cr} {c\left( 2r+1\right)} = {\left(1+r\right)}^n \\

\text{take the base } \left( 1+r\right) \text{logarithm} \\
\log_{1+r}\left(\frac {fr + c + cr} {c\left( 2r+1\right)}\right) = \frac {\log\left(\frac {fr + c + cr} {c\left( 2r+1\right)}\right)} {\log(1+r)} = n \\
$

got the same answer as galactus' TI92 so I think it looks good.

(note: also confirmed via Maxima)
 galactus Posts: 2,271, Reputation: 282 Ultra Member #7 Mar 21, 2007, 12:23 PM
Very nice, Asterisk Man. Showin' off. ;) :)

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