Question

Model - R=P(i/(1-(1+i)^(-nt))), where i =r/n, r=8% compounded monthly(n) and p=90000, find t. When R=1200
1200 = 90000(0.08/(12)/(1-(1+0.08/(12))^-(12)t Find t(year)

Answer 1

this ur model

\(\large \color{black}{\begin{align}
R=P\left(\dfrac{i}{1-(1+i)^{-nt}} \right)\hspace{.33em}\\~\\
\end{align}}\) ?

Answer 2

yes

Answer 3

Its pretty easy to solve
I guess i would take the logarithm on both ends
\[\large \color{black}{\begin{align}
90000=12000\left(\dfrac{\frac{.08}{12}}{1-(1+\frac{.08}{12})^{-12t}} \right)\hspace{.33em}\\~\\
\end{align}}\]
\[90000(1-(1+\frac{.08}{12})^{-12t}=12000(\frac{.08}{12})\]
\[(1+\frac{.08}{12})^{-12t}=-12000(\frac{.08}{12})+90000\]
Simplify and take the log on both sides

Answer 4

whoops its 1200 not 12000

Answer 5

\[(1+\frac{.08}{12})^{-12t}=-1200(\frac{.08}{12})+90000\]

Answer 6

Why did you add the 90000?

Answer 7

Its 90,000

Answer 8

and I just did basic simplifying

Answer 9

so its not?\[-90000(1+.08/12)^\left( -12t \right)=1200(.08/12)-90000\]

Answer 10

lol u r actually right -.-
Sorry i skipped out that step

Answer 11

\[(1+.08/12)^\left( -12t \right)=\frac{1200(.08/12)-90000}{-90,000}\]

Answer 12

\[\log.08/12(.08/12)^\left( -12t \right)=\log.08/12(9)\] is what i have as of now

Answer 13

hmmmm this what I have
\[\log(1+.08/12)^{-12t}=\log(\frac{1200(.08/12)-90000}{-90,000})\]

Answer 14

forgot the 1

\[\log_{1+.08/12} (1+.08/12)^\left( -12t \right)=\log_{1+.08/12}(9)\]
\[-12t=\log_{1+.08/12}(9)\]

Answer 15

I dont get why u just dont use log base 10 ... easier to work with

Answer 16

I would but, I want to cancel the 1+... so t can get separated

Answer 17

Or atleast thats what I was taught

Answer 18

Hmmm this is how I would go about it tbh
Why cant you just add 1+.08/12
Do you have a scientific calculator?

Answer 19

Actuallly do what you were taught in school :P

Answer 20

Just not sure how you got the right hand side of the equation

Answer 21

I get \[-12t=330.68...\]
\[t=-27.556...\]

Answer 22

I got 9 from solving it on a calculator

Answer 23

but, I don't think t being a negative answer, would be "correct" - in that if there is a negative year

Answer 24

Ya something doesnt look right there
Wait lemme go over ur steps

Answer 25

How did you get the 9 again?

Answer 26

Cuz I dont seem to get that answer

Answer 27

I got 9 by doing \[(1200(.08)-90000)/(-90000)\]

Answer 28

for the 12, but it was included when typed into the calculator

Answer 29

im giving up .... @precal most probably will catch my error

Answer 30

Well, I was at wrong - forgot the parenthesis

Answer 31

0.9999111111 is what you get - you forgot it was divided by a negative 90k

Answer 32

ohhhhh yup

Answer 33

It shoudl work now though

Answer 34

Good job!

Answer 35

\[\log{1+.08/12} ^\left( -12t \right)=\log{0.9999111111} \]
\[0.001806^\left( -12t \right)=\log(0.9999111111)\]

Answer 36

this would mean that its too complicated and I should just do log1+.08/12(1+.08/12)

Answer 37

\[-12t=\log{1+.08/12}(0.9999111111)\]

Answer 38

hmmm do you mean
\[-12t=\log_{1+.08/12}(.999911111)\]

Answer 39

yes

Answer 40

I keep getting a really small number for t
Makes no difference how I compute this -.-

Answer 41

same, I get 0.001115

Answer 42

\[\log_{1+(.08/12)}(1+.08/12)^{-12t}=\log_{1+(.08/12)}(\frac{1200(.08/12)-90000}{-90,000})\]
\[-12t=\log_{1+(.08/12)}(\frac{1200(.08/12)-90000}{-90,000})\]
@dan815 Can you solve for t
I keep getting an abnormally small number

Answer 43

mmk I got it to where t~8.693 by doing
\[1200=90000(i/1-(1+i)^\left( -12t \right)\]
\[-12t=\log_{1+.08/12} (.5)\]

Answer 44

From what I remember the original question was an couple owes $90000, where their i is at a rate of 8% compounded monthly. And they paid $1200 per month - how many years would it take for them to pay it off

Answer 45

I got 8.69 using TVM Solver

Answer 46

So that is def the correct answer

Answer 47

Thanks for the help.

Answer 48

How did you get the .5 though?

Answer 49

i did
\[600-1200/(-1200)\]
where 600 came from 90000(i)

Answer 50

http://www.wolframalpha.com/input/?i=r%3D0.08%2Cn%3D12%2Ci%3Dr%2Fn%2CP%3D90000%2C+1200%3DP%28i%2F%281-%281%2Bi%29^%28-n+t%29%29%29&a=i_Variable