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Thread: you invest $1000 in a savings account that pays 6% a compounded annually.

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    you invest $1000 in a savings account that pays 6% a compounded annually.

    you invest $1000 in a savings account that pays 6%/a, compounded annually.

    calculate the rate at which the amount is growing over the first

    i)2 years ii)5 years iii) 10 years


    I have a test tomorrow on exponential and logorithmic functions... These are the ONLY type of question that I struggle with? I don't even know where to start with these questions!

    any help will be appreciated
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    Re: you invest $1000 in a savings account that pays 6% a compounded annually.

    in two years ...

    $\displaystyle \frac{1000(1.06)^2 - 1000}{1000} = 0.1236$

    ... balance has increased by an average rate of 12.36%

    can you determine the growth rate for 5 & 10 years?
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    Re: you invest $1000 in a savings account that pays 6% a compounded annually.

    the book says the answer is 61.80

    I still don't really understand what formula to use with this questions. Ugh The "hard" questions are so easy for me but I just can't do these interest questions
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    Re: you invest $1000 in a savings account that pays 6% a compounded annually.

    the book answer is the average rate of change in dollars per year over the given time period, not a percentage.

    Let A be the account balance as a function of time t in years ...

    $\displaystyle A(t) = 1000(1.06)^t$

    avg rate of change for two years ...

    $\displaystyle \frac{A(2) - A(0)}{2 - 0}$

    it's just the slope of the secant line drawn between two points on the curve
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    Re: you invest $1000 in a savings account that pays 6% a compounded annually.

    Quote Originally Posted by ForeverConfused View Post
    you invest $1000 in a savings account that pays 6%/a, compounded annually.

    calculate the rate at which the amount is growing over the first

    i)2 years ii)5 years iii) 10 years
    The first one, at least can be done directly: at 6% annually, after the first year it has earned $\displaystyle 0.06(1000)= $60$ interest so is worth $1000+ 60= $1060. After the second year, it will have earned $\displaystyle 0.06(1060)= $63.60$ interest so is worth $1060+ 63.60= $1123.60.

    You could do the same thing for the third, fourth, and fifth years, then the sixth, seventh, eight, nineth, and tenth years, but as skeeter says, finding 6% of an amount, A, then adding that to A is the same thing as A+ 0.06A= (1.06)A so the second year you do the same thing to get $\displaystyle (1.06)((1.06)A= (1.06)^2A$, for the third year, $\displaystyle (1.06(1.06^2)A= (1.06^3)A$, etc. In this problem, for the second year that would be [tex](1.06)^2(1000)= (1.1236)(1000)= $112360 as before. For the fifth year that is $\displaystyle (1.06^5)(1000)$ and [tex](1.06^{10})(1000)/tex] for 10 years.


    I have a test tomorrow on exponential and logorithmic functions... These are the ONLY type of question that I struggle with? I don't even know where to start with these questions!

    any help will be appreciated
    Last edited by HallsofIvy; Nov 11th 2014 at 03:54 PM.
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