Results 1 to 12 of 12

Math Help - Solving for Number of Periods (Help with Log)

  1. #1
    Member
    Joined
    Jan 2009
    From
    Punjab, Pakistan
    Posts
    88

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member anonimnystefy's Avatar
    Joined
    Jul 2011
    Posts
    157
    Thanks
    3

    Re: Solving for Number of Periods (Help with Log)

    what is the n in the equation?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1

    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)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,965
    Thanks
    1785
    Awards
    1

    Re: Solving for Number of Periods (Help with Log)

    Quote Originally Posted by dexteronline View Post
    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)}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2009
    From
    Punjab, Pakistan
    Posts
    88

    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
    Follow Math Help Forum on Facebook and Google+

  6. #6
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1

    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}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jan 2009
    From
    Punjab, Pakistan
    Posts
    88

    Re: Solving for Number of Periods (Help with Log)

    Quote Originally Posted by e^(i*pi) View Post
    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
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Jan 2009
    From
    Punjab, Pakistan
    Posts
    88

    Re: Solving for Number of Periods (Help with Log)

    Quote Originally Posted by e^(i*pi) View Post
    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
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,162
    Thanks
    70

    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?
    Last edited by Wilmer; August 4th 2011 at 08:42 PM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Jan 2009
    From
    Punjab, Pakistan
    Posts
    88

    Re: Solving for Number of Periods (Help with Log)

    Quote Originally Posted by Wilmer View Post
    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)
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,162
    Thanks
    70

    Re: Solving for Number of Periods (Help with Log)

    Quote Originally Posted by dexteronline View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Jan 2009
    From
    Punjab, Pakistan
    Posts
    88

    Re: Solving for Number of Periods (Help with Log)

    Quote Originally Posted by Wilmer View Post
    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

    I could have never figured it out on my own
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Find number of periods from annuity formula
    Posted in the Business Math Forum
    Replies: 5
    Last Post: August 24th 2011, 11:03 AM
  2. Homomorphism and periods
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 16th 2011, 08:17 AM
  3. calculating the number of compounding periods
    Posted in the Business Math Forum
    Replies: 5
    Last Post: March 27th 2010, 12:58 PM
  4. calculating the number of compounding periods
    Posted in the Business Math Forum
    Replies: 6
    Last Post: March 16th 2010, 06:06 PM
  5. Periods
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: August 18th 2009, 01:49 AM

Search Tags


/mathhelpforum @mathhelpforum