I am doing something wrong as when i take the log of both sides of the equation then the answer I am expecting is not correct
Can someone help me solve for p in the following equation
(1+i)^p = 0.8 + 0.2(1+i)^n
Thanks
I am doing something wrong as when i take the log of both sides of the equation then the answer I am expecting is not correct
Can someone help me solve for p in the following equation
(1+i)^p = 0.8 + 0.2(1+i)^n
Thanks
Thanks you both for your help
It seems that I have done something terribly wrong with the way I have attempted to solve a business math problem
The problem I was attempting to solve has to do with finding the month in which the Outstanding Principal of the Loan is equal to 80% of the Home Value
Take this for example
Home Value is $300,000
Loan Amount is $250,000
Interest rate (annual) is 5%
Number of Years in the loan is 30
Now when I used the formula to find outstanding balance at the end of 31st month, it nearly equals 80% of the Home Value of 240,000
Outstanding balance is roughly 240,081.85
It is calculated using this formula
L [ (1+i)^n - (1+i)^p]/[(1+i)^n - 1 ]
where L is the loan amount of 250,000
i is the interest rate 5%/12
p is the period for which I seek the outstanding principal and in this case is 31
n is the number of months for loan equal to 360
Here is where I screwed up, to find p from this equation I set the equation to 0.8L
obviously it should have been set equal to 0.8HV
HV is the home value
Even when I set it equal to HV in the last 10 minutes I seem to be going no where
My attempt here is to find the month p where the outstanding principal is equal to 80% of the home value
Is there a simpler way of finding out the month where the remaining balance is equal to a certain amount in this case 80% of the home value
Sounds like a problem which is more easily solved using the compound interest formula:
Where,
* A = final amount
* P = principal amount (initial investment)
* r = annual nominal interest rate (as a decimal - it should not be in percentage)
* n = number of times the interest is compounded per year
* t = number of years
You want to find the value of t when:
Does the question say how often interested is compounded?
If it's compounded monthly:
If it's compounded continuously:
Sorry for the delayed reply
The compounding is monthly for the mortgage problem
I was testing the formula and noticed if I use 250000 the log turns out to be a negative number
Yet when I use the monthly payment amount of 1342.05, the answer is twice the number I expected
=LOG(240000/1342.05)/LOG(1+5%/12)*0.05
62.37
when I expect a value of slightly higher than 31
Can't follow what you've done...but it's wrong! This'll do it:
a = 250000
f = 240000
p = 1342.05
i = .05/12
k = 1 + i
x = (fi - p) / (ai - p)
n = log(x) / log(k) = 31.23991....
Your problem could be worded this simply:
$250000 is borrowed at rate of 5% annual compounded monthly, requiring monthly payments
of $1342.05 over 30 years. After how many months will the amount owing be $240000?
I see NO reasons to use "e".
How can a variable representing the targeted amount (f = 240000 in this case) NOT be part of equation?!
a = 250000
f = 240000
p = 1342.05
i = .05/12
k = 1 + i
I combined the FV of amount and FV of annuity formulas:
a(k^n) - p[(k^n - 1]/i = f
and solved above for n.