1. ## Intrest

If $600 is deposited in an account paying 11% compounded continously, then what amount will be in the account after 8 years? I just needed to double check if A=Pe^(rt) 2. Originally Posted by calc_help123 If$600 is deposited in an account paying 11% compounded continously, then what amount will be in the account after 8 years?
I just needed to double check if A=Pe^(rt)
Hi calc_help123,

Yes. $A=Pe^{rt}$

So, $A=600e^{(.11)(8)}\approx 1446.54$

3. Originally Posted by calc_help123
If \$600 is deposited in an account paying 11% compounded continously, then what amount will be in the account after 8 years?
I just needed to double check if A=Pe^(rt)
It is better to use the second formula below because e is only correct for large n as the binomial expansion shows: $(1+\frac{1}{n})^n \rightarrow e \: , \: n \rightarrow \infty$

$A(t) = A(0)(1+{x})^{t}$ is a better model

1. A(t) = Amount at time t
2. A(0) = Amount at time 0
3. x = amount of interest in one period
4. t = time (number of periods)

$A(t) = 600(1+\frac{11}{100})^8 = \1382.72$