Suppose $500 is invested at 6% annual interest compounded twice a year. When will it be worth $1000?

The first bit is almost correct  anon  

$$\\1000=500(1+0.03)^n\\
2=1.03^n\\
log2=log1.03^n\\
log2=nlog1.03\\
n=log2/log1.03$$

$${\frac{{log}_{10}\left({\mathtt{2}}\right)}{{log}_{10}\left({\mathtt{1.03}}\right)}} = {\mathtt{23.449\: \!772\: \!250\: \!437\: \!735\: \!9}}$$

It will be worth 1000 in 24 lots of 6 months = 12 years

I guess I have to dig out my compound interest formula :P

$${\mathtt{A}} = {{P}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{r}}\right)}}^{\,{\mathtt{n}}}$$

Where A is the amount,

P is the original value and,

r is the rate and

n is the number of compound periods.

So, lets fill out the stuff we know.

$${\mathtt{1\,000}} = {\mathtt{500}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.06}}\right)}^{{\mathtt{n}}}$$

This might not be correct but I believe it would be rearranged as so,

$${\mathtt{n}} = {\sqrt{{\mathtt{500}}{\mathtt{\,\times\,}}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.06}}\right)}} \Rightarrow {\mathtt{n}} = {\mathtt{23.021\: \!728\: \!866\: \!442\: \!676\: \!4}}$$

Thats how many compound periods it would take.

To find the answer in years divide by the amount of compound periods in a year.

Good luck!