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Math Help - [SOLVED] simplify Algebra Problem with exponents

  1. #1
    gnom
    Guest

    [SOLVED] simplify Algebra Problem with exponents

    how do you solve for R, rate in this equation
    please provide a "check" that shows how to solve the problem

    Formula
    I = G/(1+r) + G/(1+r)^2 + G/(1+r)^3
    Given
    I=100,000
    G=40211 (which is from the fact that 120634/3 = 40211)
    Solve for R
    100,000 = 40211/(1+r) + 40211/(1+r)^2 + P/(1+r)^3

    About question
    this problem is supposed show how a stream of 120634 divided into 3 equal payments of 40211 can yield the same rate of return of 100,000 with respect to the rate of return. we are solving for the rate of return.

    120634/3 = 40211 = G

    I know the answer for r, but I need to understand the algebraic concepts that arrive at the conclusion. (the answer is R = .10, this was done by inputing values)

    I = 100,000 = 40211/(1.1)^1 + 40211/(1.1)^2 + 40211/(1.1)^3

    I need to know how to solve for the rate r which is equal to 0.1.
    (1+r) = 1.1

    I'm having difficulty solving this problem because I do not know how to simplify the exponents for the term involving the variable (1+R).

    This problem is difficult because the term 1+r is repeated three times, and each time to a higher consective exponent. I do not know how to simplify such a complex term.

    I do know how to simplify the exponent we use log, but having three terms complicates things for me, so I need some help.

    This is a algebra problem involving exponents, but here we want to solve for the rate, R, by showing a check.

    Work Done So far (has errors)
    A = p/(1+r) + p(1+r)^2 + p(1+r)^3120634/3 = 40211
    100000 = 40211/(1+r) + 40211/(1+r)^2 + 40211/(1+r)^3
    ln 100000/40211 = ln40211/ln(1+r) + ln40211/ln(1+r)^2 + ln40211/ln(1+r)^3
    5 = ln40211/ln(1+r) + ln40211/ln(1+r)^2 + ln40211/ln(1+r)^3
    5 = ln40211/ln(1+r) + 40211/2ln(1+r) + ln40211/3ln(1+r)
    5/ln(1+r) = ln40211/1 + ln40211/2 + ln40211/3
    1/ln(1+r) = ln40211/5 + ln40211/10 + ln40211/15
    ln(1+r) = 5/ln(40211) + 10/ln(40211) + 15/ln(40211)
    ln(1+r) = 2.82968
    1+r = 16.94
    r = 15.94

    15.94 is not correct. the answer should be 0.10 resulting in 10%

    If someone can help point out where my approach was wrong.

    Another approach is considering that 1/(1+r) = 1+r^-1, but I arrived at the same answers. I think there is a error in my use of the Ln function

    As you can see I need some clarification on the logarthim rules for algebra

    If someone can provide a check, That would be very much appreciated


    PS
    I need to know how to solve for this problem, because on the test they will not tell me what the rate is equal to, so this is similar to the compound interest formula, but here we solve for rate, and rate is complicated by the fact that it is in the denominator, it is repeated, and it is taken the consective higher exponent.
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  2. #2
    MHF Contributor
    Joined
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    There are general solutions for 3rd and 4th degree equations, but it is highly doubtful that you would want to use them in a practical application. Unless it's quadritic or linear, finding the interest rate is a job for numerical methods.

    Did you ever notice that your calculator solves immediately for Term, Present Value, Future Value, or Payment Amount, but takes sometimes quite a while to find a missing interest Rate? This is why. Direct solutions are available for everything except the interest rate.
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