Future value formula is:

$A=P \cdot (1+\frac{r}{m})^{m \cdot t}$


  • $A$ 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?

  • $\begingroup$ Let $m=nr$. Then allow $n\rightarrow\infty$. Then observe that $r=k$ $\endgroup$ – CogitoErgoCogitoSum Jul 18 at 19:13

If I understand the notation in your question, I see a couple of items that seem they should be addressed:

  1. 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.

  2. 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.

| cite | improve this answer | |

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.

| cite | improve this answer | |

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 $(1+\frac rm)^{mt}$.

| cite | improve this answer | |
  • $\begingroup$ How did you calculate and get ≈43123 in the first answer? Because I'm calculating again and again but it results ≈ 2.47144x10^12? $\endgroup$ – Nay Sie Jul 18 at 10:25
  • $\begingroup$ @NaySie I fixed a typo. The exponent is $kt$ with $k=.0812$ and $t=20$ in years. $\endgroup$ – grand_chat Jul 18 at 22:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.