Question 428877
the formula is {{{A = P(1.03)^t}}}.
Doubling time:
{{{2P =  P(1.03)^t}}}
{{{2 = (1.03)^t}}
{{{ln2 = (ln1.03)t}}}
{{{t = (ln2)/ln(1.03) = 23.45}}} years, to two decimal places.  You're answer is pretty close.

In 5 years, {{{A = P(1.03)^5 = 1.159274P}}}, so the indicator will increase by approximately 16%.