Solve in terms of n

tboss · #1 Mar 19, 2007, 03:20 PM

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 · Mar 20, 2007, 12:45 AM

Hi guys, someone please throw me a lifeline!!!

tboss · Mar 20, 2007, 01:01 PM

I've heard that MathCad can be used to solve this... I don't have MathCad. Can someone try it out?

tboss · Mar 20, 2007, 01:02 PM

I've heard that this can be done with MathCad. I don't have MathCad. Can someone please help try? BTW, This is not homework!!!

asterisk_man · Mar 21, 2007, 10:36 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)



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