Why do two exponential growth functions give 2 different answers? We had this question on a test and it's kinda puzzling for me.

A private high school charges $57,000 for tuition, but this figure is expected to rise 10% per year. What will tuition be in 3 years?
a) $68,970
b) $75,867.00
c) $83,453.70
d) $76,942,00 

Using the formula $y(t) = a \times (1+k)^t$, where $a$ is initial tuition, $k$ is rate and $t$ is time, we get 
$$57,000 \times (1+0.1)^3=\$75,867.00$$
which is also the correct the answer is actually (choice B). However, I used the exponential growth formula 
$y(t)= a × e^{kt}$, which gives
$$57,000 \times e^{0.1 \times 3} = \$76,942,00 $$
which is choice D!
I can't really distinguish between the 2 equations and why B is correct while D isn't.  
 A: It can be the same but you need to alter the rate, notice that
$$
57000 \cdot e^{\ln(1 + 0.1) t} = 57000 \cdot (e^{\ln(1 + 0.1)})^t = 57000 \cdot (1 + 0.1)^t
$$
So if you wanted to use the model
$$
a \cdot e^{k t}
$$
then you must set $k = \ln(1 + 0.1)$. 
A: Observe that the function you used to calculate the answer $b$ is:
$$y(t)=a(1+\kappa)^t$$
It then follows that: $$a(1+\kappa)^t=ae^{kt}\implies1+\kappa=e^k\implies k=\ln(1+\kappa)$$
A: The reason is that the value $r$ in
$$A(t) = Pe^{rt}$$
is not the "percentage growth rate per year", but rather something different, and hence you cannot exchange it with the $r_\mathrm{pct}$ in
$$A(t) = P (1 + r_\mathrm{pct})^t$$
which is that percentage growth rate. To understand it better, I think it may perhaps be more insightful to write the formula in terms of the reciprocal of $r$, which is the time constant $\tau$:
$$\tau := \frac{1}{r}$$
so
$$A(t) = Pe^{t/\tau}$$
You can see then, by taking $t = 1\ \tau$, $t = 2\ \tau$, etc. that $\tau$ is then the $e$-folding time for the amount: it is the unit of time that has to elapse in order for the amount $A$ to grow up by one factor of $e$ from the amount it was one such unit of time prior. Hence, the $r$ in the equation using $e$ is perhaps better thought of as the "number of $e$-foldings per year", which is quite different from the percent growth per year. One $e$-folding in a year, i.e. $r = 1$, is about 171.8% growth! The two are related, of course, by
$$1 + r_\mathrm{pct} = e^r$$
so $r = \ln(1 + r_\mathrm{pct})$, and so for your problem with a 10% growth rate, i.e. $r_\mathrm{pct} = 10\% = 0.10$, $r \approx 0.0953$ e-foldings per year. Indeed, you should see that, using the logic by which you arrived at answer B), after 10 years, the principal will have grown by a factor a bit less than $e$. To get answer B) using the $e$-based formula, you need the value $r$ computed above, i.e. $\ln 1.10$.
