Continuing from zacismyname's analysis:
Take logs of both sides:
log(1.15n) = log(20/9)
Because log(an) = n*log(a) we can write
n*log(1.15) = log(20/9)
so
n = log(20/9)/log(1.15)
$${\mathtt{n}} = {\frac{{log}_{10}\left({\frac{{\mathtt{20}}}{{\mathtt{9}}}}\right)}{{log}_{10}\left({\mathtt{1.15}}\right)}} \Rightarrow {\mathtt{n}} = {\mathtt{5.713\: \!341\: \!433\: \!638\: \!366\: \!8}}$$
n ≈ 5.7 years
.
Trout Population = starting population*(1 + rate)^number of years
$$P = SP\times{(1+0.15)^n$$
$$500 = 225\times{(1+0.15)^n$$
$$1.15^n = \frac{20}{9}$$
Now I'm not sure how you find n. It is somewhere between 5 and 6.
Continuing from zacismyname's analysis:
Take logs of both sides:
log(1.15n) = log(20/9)
Because log(an) = n*log(a) we can write
n*log(1.15) = log(20/9)
so
n = log(20/9)/log(1.15)
$${\mathtt{n}} = {\frac{{log}_{10}\left({\frac{{\mathtt{20}}}{{\mathtt{9}}}}\right)}{{log}_{10}\left({\mathtt{1.15}}\right)}} \Rightarrow {\mathtt{n}} = {\mathtt{5.713\: \!341\: \!433\: \!638\: \!366\: \!8}}$$
n ≈ 5.7 years
.