anonymous
  • anonymous
Algebrea, solve for I. Please show me all the steps? PV = FV/(1 + I)^N
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[PV=FV(1+L)^n\] like that?
anonymous
  • anonymous
the letter I (eye) The answer is I =(fv/pv) ^1/N -1..... but I am getting stuck after I cross multiply
anonymous
  • anonymous
ok ok don't do that divide by \(FV\) first and get \[\frac{P}{F}=(1+I)^n\]

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anonymous
  • anonymous
then take the nth root to get \[1+I=\sqrt[n]{\frac{P}{F}}\]
anonymous
  • anonymous
what happened to the latex lol
anonymous
  • anonymous
then subtract 1
anonymous
  • anonymous
\[\begin{align*}PV&=\frac{FV}{(1+I)^N}\\ (1+I)^N&=\frac{FV}{PV}\\ \ln(1+I)^N&=\ln\frac{FV}{PV}\\ N\ln(1+I)&=\ln\frac{FV}{PV}\\ \ln(1+I)&=\frac{1}{N}\ln\frac{FV}{PV}\\ \ln(1+I)&=\ln\left(\frac{FV}{PV}\right)^{1/N}\\ e^{\ln(1+I)}&=e^{\ln\left(\frac{FV}{PV}\right)^{1/N}}\\ 1+I&=\left(\frac{FV}{PV}\right)^{1/N}\\ I&=\cdots \end{align*}\]
anonymous
  • anonymous
if you prefer you can write \[\sqrt[n]{1+I}=(1+I)^{\frac{1}{n}}\]
anonymous
  • anonymous
@SithsAndGiggles at what point to you cancel the \(V\) ?
anonymous
  • anonymous
@satellite73 I took \(PV\) and \(FV\) to stand for present/future value.
anonymous
  • anonymous
oooh ok one term i see now i have a second question why take the log and then exponentiate?
anonymous
  • anonymous
Yeah, I was gonna say your method was quicker... Oh well, another way to get the same result.
anonymous
  • anonymous
\[x^2=25\] \[\ln(x^2)=\ln(25)\] \[2\ln(x)=\ln(25)\] ...
anonymous
  • anonymous
I suppose this way you get rid of the negative root...
anonymous
  • anonymous
oooh i see
anonymous
  • anonymous
at least you didn't write "present value over future value" is "present over future" like i did lol
anonymous
  • anonymous
the instructor got the answer without using steps 3 - 5
anonymous
  • anonymous
@Iron, right, refer to satellite's steps. My method is a bit roundabout.
anonymous
  • anonymous
what is the nth root?
anonymous
  • anonymous
Given something like \[x^n=10\] you'd solve for \(x\) by taking the nth root of both sides: \[\left(x^n\right)^{1/n}=10^{1/n}\\ x=10^{1/n}\]
anonymous
  • anonymous
So in the second step, \[(1+I)^N=\frac{FV}{PV}\] taking the nth root of both sides gives you \[\left((1+I)^N\right)^{1/N}=\left(\frac{FV}{PV}\right)^{1/N}\\ 1+I=\left(\frac{FV}{PV}\right)^{1/N}\]
anonymous
  • anonymous
i'm dumb. i still don't get it
anonymous
  • anonymous
For instance, when you solve the equation \(x^2=4\), how do you go about it?
anonymous
  • anonymous
to be honest i dont remember

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