Investment interest problem

"An investment of $1000 earns $177.23 in interest in two years. If the interest is compounded annually, find the annual interest rate. Round to the nearest tenth of a percent.)

Using the formula $\displaystyle P = A(1 + i)^2$, I solved it thus:

$\displaystyle 1177.23 = 1000(1 + i)^2$

$\displaystyle 1.17723 = (1 + i)^2$

$\displaystyle log 1.17723 = 2 log (1 + i)$

$\displaystyle log 1.17723 = 2 log 1 + 2 log i$

$\displaystyle \frac {log 1.17723}{2} = log i$

$\displaystyle .0354306605 = log i$

$\displaystyle 10^{.0354306605} = i$

$\displaystyle i \approx 1.085$

As it turns out, my answer is correct if you remove the integer portion and multiply by 100, i.e. 8.5%. However, the answer I get is actually 108.5%. Where did I go wrong?

The book gives a hint... "The solution will require taking the common logarithm of each side of the equation and taking the common antilogaritm of each side of the equation."... but I'm not quite understanding how to do that.

Thanks in advance for your help!