# Solving for Number of Periods (Help with Log)

• Aug 4th 2011, 11:21 AM
dexteronline
Solving for Number of Periods (Help with Log)
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
• Aug 4th 2011, 11:33 AM
anonimnystefy
Re: Solving for Number of Periods (Help with Log)
what is the n in the equation?
• Aug 4th 2011, 11:38 AM
e^(i*pi)
Re: Solving for Number of Periods (Help with Log)
If you take the natural log of both sides you get: $p \ln (1+i) = \ln(0.8+0.2(1+i)^n)$. This is because of the law that says $\ln(a^b) = b\ln |a|$

However you will not be able to simplify the RHS since $\ln(a+b) \neq \ln(a)+\ln(b)$
• Aug 4th 2011, 11:42 AM
Plato
Re: Solving for Number of Periods (Help with Log)
Quote:

Originally Posted by dexteronline
Can someone help me solve for p in the following equation
(1+i)^p = 0.8 + 0.2(1+i)^n

Because this is in business math, I assume that $i$ is for interest.

So $p=\frac{\log\left[0.8+0.2(1+i)^n\right]}{\log(1+i)}$
• Aug 4th 2011, 11:55 AM
dexteronline
Re: Solving for Number of Periods (Help with Log)
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
• Aug 4th 2011, 12:04 PM
e^(i*pi)
Re: Solving for Number of Periods (Help with Log)
Sounds like a problem which is more easily solved using the compound interest formula: $A = P \left(1+\dfrac{r}{n}\right)^{rt}$

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: $A = 0.8 \times 300,000 = 240000$

Does the question say how often interested is compounded?

If it's compounded monthly: $240000 = 250000\left(1+\dfrac{0.05}{12}\right)^{0.05t}$

If it's compounded continuously: $240000 = 250000 \cdot e^{0.05t}$
• Aug 4th 2011, 12:34 PM
dexteronline
Re: Solving for Number of Periods (Help with Log)
Quote:

Originally Posted by e^(i*pi)
Sounds like a problem which is more easily solved using the compound interest formula: $A = P \left(1+\dfrac{r}{n}\right)^{rt}$

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: $A = 0.8 \times 300,000 = 240000$

Does the question say how often interested is compounded?

If it's compounded monthly: $240000 = 250000\left(1+\dfrac{0.05}{12}\right)^{0.05t}$

If it's compounded continuously: $240000 = 250000 \cdot e^{0.05t}$

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
• Aug 4th 2011, 03:15 PM
dexteronline
Re: Solving for Number of Periods (Help with Log)
Quote:

Originally Posted by e^(i*pi)
Sounds like a problem which is more easily solved using the compound interest formula: $A = P \left(1+\dfrac{r}{n}\right)^{rt}$

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: $A = 0.8 \times 300,000 = 240000$

Does the question say how often interested is compounded?

If it's compounded monthly: $240000 = 250000\left(1+\dfrac{0.05}{12}\right)^{0.05t}$

If it's compounded continuously: $240000 = 250000 \cdot e^{0.05t}$

I had to resort to the Newton Raphson method to find the month in which the outstanding principal equals a given amount

I am sure there is some sort of a simpler formula that gets the answer, would have saved me the trouble from programming the NR method
• Aug 4th 2011, 07:32 PM
Wilmer
Re: Solving for Number of Periods (Help with Log)
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? • Aug 5th 2011, 12:01 AM dexteronline Re: Solving for Number of Periods (Help with Log) Quote: Originally Posted by Wilmer 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?

Many Thanks Wilmer, This is it a nice simple formula to find time period

Did you use the same annuity formula as mentioned by e^(i*pi) since I see a variable F which was not there in formula provided by e^(i*pi)
• Aug 5th 2011, 04:18 AM
Wilmer
Re: Solving for Number of Periods (Help with Log)
Quote:

Originally Posted by dexteronline
Did you use the same annuity formula as mentioned by e^(i*pi) since I see a variable F which was not there in formula provided by e^(i*pi)

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.
• Aug 5th 2011, 05:07 AM
dexteronline
Re: Solving for Number of Periods (Help with Log)
Quote:

Originally Posted by Wilmer
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.

You are a genius (Clapping)

I could have never figured it out on my own