The time required to double the amount of an investment at an interest rate

I am just going to do the first one for you.    (Completed after Alan's comment below)

http://cs.selu.edu/~rbyrd/math/continuous/

$$A=Pe^{rt}$$

r=5%=0.05

A=2P

$$\begin{array}{rll}
2P&=&Pe^{0.05t}\\\\
2&=&e^{0.05t}\\\\
ln2&=&lne^{0.05t}\\\\
ln2&=&0.05t\\\\
0.05t&=&ln2\\\\
t&=&\frac{ln2}{0.05}\\\\
\end{array}$$

$${\frac{{ln}{\left({\mathtt{2}}\right)}}{{\mathtt{0.05}}}} = {\mathtt{13.862\: \!943\: \!611\: \!198\: \!906\: \!2}}$$

13.86 time units needed to double the investment.

Melody, if you put A = 2P (investment doubled) then take logs you get ln(2) = r*t or t = ln(2)/r.

Thanks Alan,

okay I did the 2P bit (before I deleted it) but I was confused because the ln2 was written in the question.

But what you have said makes sense.  I'll go back and finish it.