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Mathematics 21 Online
OpenStudy (anonymous):

Algebrea, solve for I. Please show me all the steps? PV = FV/(1 + I)^N

OpenStudy (anonymous):

\[PV=FV(1+L)^n\] like that?

OpenStudy (anonymous):

the letter I (eye) The answer is I =(fv/pv) ^1/N -1..... but I am getting stuck after I cross multiply

OpenStudy (anonymous):

ok ok don't do that divide by \(FV\) first and get \[\frac{P}{F}=(1+I)^n\]

OpenStudy (anonymous):

then take the nth root to get \[1+I=\sqrt[n]{\frac{P}{F}}\]

OpenStudy (anonymous):

what happened to the latex lol

OpenStudy (anonymous):

then subtract 1

OpenStudy (anonymous):

\[\begin{align*}PV&=\frac{FV}{(1+I)^N}\\ (1+I)^N&=\frac{FV}{PV}\\ \ln(1+I)^N&=\ln\frac{FV}{PV}\\ N\ln(1+I)&=\ln\frac{FV}{PV}\\ \ln(1+I)&=\frac{1}{N}\ln\frac{FV}{PV}\\ \ln(1+I)&=\ln\left(\frac{FV}{PV}\right)^{1/N}\\ e^{\ln(1+I)}&=e^{\ln\left(\frac{FV}{PV}\right)^{1/N}}\\ 1+I&=\left(\frac{FV}{PV}\right)^{1/N}\\ I&=\cdots \end{align*}\]

OpenStudy (anonymous):

if you prefer you can write \[\sqrt[n]{1+I}=(1+I)^{\frac{1}{n}}\]

OpenStudy (anonymous):

@SithsAndGiggles at what point to you cancel the \(V\) ?

OpenStudy (anonymous):

@satellite73 I took \(PV\) and \(FV\) to stand for present/future value.

OpenStudy (anonymous):

oooh ok one term i see now i have a second question why take the log and then exponentiate?

OpenStudy (anonymous):

Yeah, I was gonna say your method was quicker... Oh well, another way to get the same result.

OpenStudy (anonymous):

\[x^2=25\] \[\ln(x^2)=\ln(25)\] \[2\ln(x)=\ln(25)\] ...

OpenStudy (anonymous):

I suppose this way you get rid of the negative root...

OpenStudy (anonymous):

oooh i see

OpenStudy (anonymous):

at least you didn't write "present value over future value" is "present over future" like i did lol

OpenStudy (anonymous):

the instructor got the answer without using steps 3 - 5

OpenStudy (anonymous):

@Iron, right, refer to satellite's steps. My method is a bit roundabout.

OpenStudy (anonymous):

what is the nth root?

OpenStudy (anonymous):

Given something like \[x^n=10\] you'd solve for \(x\) by taking the nth root of both sides: \[\left(x^n\right)^{1/n}=10^{1/n}\\ x=10^{1/n}\]

OpenStudy (anonymous):

So in the second step, \[(1+I)^N=\frac{FV}{PV}\] taking the nth root of both sides gives you \[\left((1+I)^N\right)^{1/N}=\left(\frac{FV}{PV}\right)^{1/N}\\ 1+I=\left(\frac{FV}{PV}\right)^{1/N}\]

OpenStudy (anonymous):

i'm dumb. i still don't get it

OpenStudy (anonymous):

For instance, when you solve the equation \(x^2=4\), how do you go about it?

OpenStudy (anonymous):

to be honest i dont remember

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