Interest rate compounded continuously (pre-calculus) (a) Suppose you deposit $P$ dollars into a bank that pays an interest rate $r$ compounded continuously. How long does it take to double your original deposit $P$.
(b) Suppose you deposit $P$ dollars into a bank that compounds interest continuously. What is the interest rate $r$ that doubles your original investment $P$ after the first year.
The equation for continuous compound interest is $A = Pe^{rt}$ where $P$ = principal value, $r$ = interest rate per year, $t$ = time in years, $A$ = amount, and $e$ = the mathematical constant $e$
I've been working on this question for hours now, but I don't know if the answers I got are correct so I would appreciate some confirmation
The answer I got for a. is $t = (\ln2A/P)/r$
The answer I got for b. is $r = (\ln2A/P)/t$

EDIT1:
a. Time to double the principal P:
The results  you obtained after infinitesimal time compounding in exponential form is correct. Now after cancelling $(A,P)$ magnitudes its quotient is 2,
$$e^{rt}=\dfrac{A}{P} =2$$
or simply (you need to plug in A/P value!)
$$ rt =\ln2,\, t=\dfrac {\ln 2}{r}  $$ 
b. Humongous interest rate to double the principal in one year itself
Now doubling the principal in one year needs an exorbitant charging (so least it deserves is representation with a capital $R$!)
$$ e^{\,1\cdot R}= \dfrac{A}{P}= 2 \rightarrow R = ln\,2 \approx 0.69315 $$
so the rate  of interest would be a whopping
$$ R \approx 69.315\text{ % }.$$
