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 2. ## 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?

3. ## 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 4. ## 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

5. ## Re: you invest $1000 in a savings account that pays 6% a compounded annually. Originally Posted by ForeverConfused 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