An interesting situation

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Check out this dialogue between two calculus students (based on a true story):

Devyn: Riley, I’ve been thinking about the interest on my bank account.
Riley: So, like the compound interest formula, $A = P\left ( 1 + \frac {r}{n} \right )^{nt}?$
Devyn: Yes! To make it easier to work with, suppose I deposit just $1, so $P=1$ .
Riley: A $1 bank account balance? I can imagine that!
Devyn: If the interest is compounded monthly, $n=12$ , and if the deposit is at $3\%$ interest, then $r = 0.03$ .
Riley: Since $\dfrac {0.03}{12} = .0025$ , that brings you to $A = \left ( 1.0025 \right )^{12t}.$
Devyn: My question is: How long does it take the account to double in value? How long until the account balance is $2?
Riley: Can’t we just setup $\left (1.0025\right )^{12t} = 2 ?$
Devyn: Of course, but how to solve that for $t$ ? The variable is up in an exponent.
Riley: Hmmmm. I’m not sure…

What kind of operation will allow us to solve for $t$ ?

Take the square root of both sides. Take the cosecent of both sides. Raise both sides to the $1.0025$ -th power. Take a logarithm of both sides.

Take the compound interest formula above $\displaystyle A = P\left ( 1 + \frac {r}{n}\right )^{nt}$ with $P = 1$ , $r = 1$ , and $t = 1$ : $A = \left ( 1 + \frac {1}{n} \right )^{n}.$ Plug in larger and larger values of $n$ and see what happens to the values of $A$ .

$\dfrac {1}{n}$ tends to $0$ , so $\displaystyle \lim _{n \to \infty } \left (1 + \frac {1}{n} \right )^n = 1$ . The exponent is getting bigger and bigger, so $\displaystyle \lim _{n \to \infty } \left (1 + \frac {1}{n} \right )^n = \infty$ . $\displaystyle \lim _{n\to \infty } \left ( 1 + \frac {1}{n} \right )^n = 3$ . $\displaystyle \lim _{n\to \infty } \left ( 1 + \frac {1}{n} \right )^n = 2.71828\ldots$ .