Question 953769


there are 2 basic equations that i know of that can be used.


the first is f = p * (1 + r)^t


the second is f = p * e^(rt)


using the first equation of f = p * (1 + r)^t .....


f = 100
p = 50
r = what you want to find.
t = 2 months.


equation becomes:


100 = 50 * (1 + r)^2


divide both sides of this equation by 50 and then take the square root of both sides of the equation to get:


100/50 = (1 + r)^2 which becomes:
2 = (1 + r)^2 which becomes:
sqrt(2) = 1 + r


your equation started off as:


100 = 50 * (1 + r)^2


and it ended off as:


100 = 50 * (sqrt)^2


the general equation becomes:


y = 50 * sqrt(2)^t


you keep the present value at 50.
you replace 100 with y.
you replace 2 with t.


that generalizes the equation for any values of t.


the value of t = 2 was just used to find the rate per time period.
t is the number of time periods.
in this case the number of time periods are expressed in months.
the generalized equation of y = sqrt(2)^t finds the value of y for any value of t.


the alternate method is to use f = p * e^(rt)


in your situation, when you make f = 100 and p = 50 and t = 2, this equation becomes:


100 = 50 * e^(2r)


divide both sides of the equation by 50 to get:


2 = e^(2r)


take the natural log of both sides of the equation to get:


ln(2) = ln(e^(2r))


since ln(e^(2r)) is equal to 2r * ln(e) and since ln(e) is equal to 1, your equation becomes:


ln(2) = 2r


solve for r to get r = ln(2)/2


replace r in your original equation with ln(2)/2 and you get:


100 = 50 * e^(ln(2)/2 * 2)


replace 100 with y and replace t = 2 with 5 and your equation becomes:


y = 50 * e^(ln(2)/2 * t)


well, that's not exactly the equation you got.


let's test to see if all 3 equations provide the same answer.


we'll let t = 15.


first equation of y = 50 * (sqrt(2)^t becomes y = 50 * (sqrt(2)^15 which becomes:


y = 9050.966799


second equation of y = 50 * e^(ln(2)/2 * t becomes y = 50 * e^(ln(2)/2 * 15) which becomes:


y = 9050.966799


third equation of y = e^(ln(2)/2 * t + ln(50)) becomes y = e^(ln(2)/2 * 15 + ln(50)) which becomes:


9050.966799


the third equation is your equation.


i've never seen it done that way, but it works.


let me see if i can understand why.


i got y = 50 * e^(ln(2)/2 * t)


you got y = e^(ln(2)/2 * t + ln(50))


I believe that e^(ln(50)) is equal to 50, so we'll replace 50 in my equation with e^(ln(50)) to get:


y = e^(ln(50)) * e^(ln(2)/2 * t) which results in:


y = e^(ln(50) + ln(2)/2 * t) which is identical to your equation.


i can see that they're equivalent but i don't see how you arrived at your equation.


if you get a chance, let me know.


in the meantime, i hope the explanation of how y = 50 * sqrt(2)^t is satisfactory for you.