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Math Help - Compound Increase

  1. #1
    Avi
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    Compound Increase

    Hi I am studying after a few years break and have forgotten a lot of high school maths. Could someone help me with this question:
    The rate of fossil fuel emission is 40 ton per year presently. Another 565 tons of emission will result in 2degree Celsius temperature increase. If the rate of fossil fuel emission goes up by 1% per year, how long will it take to reach 2 degree Celsius increase.
    Remember, the increase is compounded each year by one percent. Thus, I would think it is something like this:

    (40 x 1.01) + (40 x 1.01 x 1.01) + (40 x 1.01 x 1.01 x 1.01)+......... = 565

    How can this be solved??
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  2. #2
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    Re: Compound Increase

    Hello, Avi!

    40(1.01) + 40(1.01)^2 + 40(1.01)^3 + \hdots + 40(1.01)^n \:=\:565

    \text{We have: }\:40(1.01)\underbrace{\big[1 + 1.01 + (1.01)^2 + (1.01)^3 + \hdots + (1.01)^{n-1}\big]}_{\text{geometric series}} \;=\;565

    The geometric series has: first term a = 1, common ratio r = 1.01, and n terms.

    . . Its sum is: . 1\cdot\frac{1.01^n-1}{1.01-1} \:=\:\frac{1.01^n-1}{0.01} \:=\:100(1.01^n-1)


    \text{We have: }\:40(1.01)\cdot100(1.01^n-1) \;=\;565 \quad\Rightarrow\quad 4040(1.01^n-1) \;=\;565

    . . . . . . . 1.01^n-1 \;=\;\tfrac{565}{4040} \;=\;\tfrac{113}{808} \quad\Rightarrow\quad 1.01^n \;=\;\tfrac{113}{808}+1 \;=\;\tfrac{921}{808}

    Take logs: . \ln\left(1.01^n\right) \;=\;\ln\left(\tfrac{921}{808}\right) \quad\Rightarrow\quad n\ln(1.01) \;=\;\ln\left(\tfrac{921}{808}\right)

    Therefore: . n \;=\;\frac{\ln(\frac{921}{808})}{\ln(1.01)} \;=\; 13.15513822
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