How to convert $P \cdot (1+\frac{r}{m})^{m \cdot t}$ to $P_0 \cdot e^{k \cdot t}$? Future value formula is:
$A=P \cdot (1+\frac{r}{m})^{m \cdot t}$
where,

*

*$A$ is resulting amount

*$r$ is annual interest

*$P$ is present value

*$n$ is number of compound periods per year

*$t$ is time (in years)

And, exponential growth function is:
$P(t) = P_0 \cdot e^{k \cdot t}$
The question is:
A retirement account is opened with an initial deposit of $8,500 and earns 8.12% interest compounded monthly. What will the account be worth in 20 years? What if the deposit was calculated using simple interest? Could you see the situation in a graph? From what point one is better than the other?

So to calculate the account worth in 20 years with exponential growth formula:
$P_0$ is $8,500$ and $k$ is $0.812$, months in 20 years is $P(240)$ and so:
for the account worth in 20 years is:
$P(240)=8500 \cdot e^{0.812 \cdot 240} = 3.67052\dots E88$
After calculating with future value formula, the answer is different:
$A = 8500 \cdot (1+\frac{0.812 \cdot 12}{12})^{12 \cdot 20} = 7.71588\dots E65 $
I see different values when I calculate with exponential growth functions and future value formula.
How to achieve this calculation correctly with exponential growth function? Is it possible?
 A: If I understand the notation in your question, I see a couple of items that seem they should be addressed:

*

*The annual interest rate is $8.12$% which is $r=0.0812$, not $r=0.812$.  Also, usually when interest rates are given, they generally refer to "annual" or "yearly" rates.


*In the future value calculation, you don't need to multiply $0.0812$ by $12$, because this is already the annual interest rate.
With the above two modifications, one has:
$$
A=8500\left(1+\frac{0.0812}{12}\right)^{12 \cdot 20}=42888.18
$$
I believe to compute the "simple interest" values, one uses the formula:
$$
A_{simple}=8500\left(1+0.0812 \cdot 20\right)=22304
$$
More details here:  https://en.wikipedia.org/wiki/Compound_interest#Calculation
I hope this helps.
A: We can use the follwoing approximation. For large $m$ we have $$\left(1+\frac{x}m \right)^{n\cdot m}\approx e^{x\cdot n}$$
With $x=0.0812, m=12$ and $n=20$ the terms are
$$8500\cdot \left(1+\frac{0.0812}{12} \right)^{12 \cdot 20}=42,888.18...$$
$$8500\cdot e^{0.0812\cdot 20}=43,123.4...$$
So the approximation in this case is not so good since $m$ is not large enough. But it goes in the right direction. The larger $m$ is, the closer are the results.
A: There are some errors in your calculation. First, the value of $k$ is $0.0812$, not $0.812$. Plug this into the exponential growth formula to get
$$P(240)=8500\cdot e^{0.0812\cdot 20}\approx 43123,$$
a more reasonable value than $3.67\times 10^{88}$.
Second, you substituted into the future value formula incorrectly.  Using $r=.0812$ you should get
$$
A=P_0\left(1+\frac rm\right)^{mt}=8500\cdot\left(1 + \frac {.0812}{12}\right)^{12 \cdot 20}\approx 42888.
$$
Note that these values are close but not exactly the same, because the exponential growth formula $e^{rt}$ is only an approximation to the future value formula $(1+\frac rm)^{mt}$.
