$x$
by a small annual $r$
percent cumulative interest rate approximates the number of years needed to double your investment with a bank. Find $x$
. Hint: The word “small” may be important.

 For which real values of
$a$
does$(1+a)^x=e$
yield real values of$x$
?  If the number of bacteria in a sample doubles every hour, how many are there after 10 hours, if the initial population was 1?
 Using Taylor expansion about 0 (Maclaurin), find
$\sin(0.1)$
correct to 2 decimal places.
 For which real values of

For
$r$
percent cumulative interest every year, how much money will you have after$n$
years? 
If the amount of money doubles after
$n$
years, find an expression relating$n$
to$r$
. 
Can you use the fact that
$r$
is small and the Taylor series for$\ln(1+x)$
to simplify your expression?

[Trivia: Luca Pacioli in 1494 said “72” (for slightly larger
$r$
), and this result was apparently already well known at that time.]The question asks us to find
$x$
, which when divided by$r$
approximates the number of years needed to double the investment. This translates to$\frac{x}{r} = n.$
Let the initial investment amount be$a_0$
, and the amount in the bank after$n$
years be$a_n$
. Each year, the amount increases by$r$
percent, which gives us the recurrence$a_{n+1} = a_n\big(1+\frac{r}{100}\big).$
Unwinding this recursion, one finds that$a_n = a_0 \big(1+\frac{r}{100}\big)^n.$
We wish to find
$n$
such that$a_n = 2a_0.$
Substitute above to obtain$2 = \big(1+\frac{r}{100}\big)^n,$
and apply log to get$n = \frac{\ln 2}{\ln(1+\frac{r}{100})}.$
Since$r$
is very small, we can approximate the denominator by using only the first term of the Taylor expansion of$\ln(1+\frac{r}{100}) \approx \frac{r}{100},$
which gives us$n = \frac{100 \ln 2} {r}.$
We now substitute into the previous equation to get$x = 100 \ln 2 \approx 69.3.$
The Taylor expansion of
$\ln(1+x)$
around$x=0$
is$\ln(1+x)=x\frac{x^2}{2}+\frac{x^3}{3}\cdots$
.