Converting from base $2$ to base 3 
Transform the binary expansion $y = 0.110110110\ldots$ into a ternary expansion.

We are given that $y = 0.110110110\ldots_2$ and thus $1000_2y = 110_2+y \implies y = \frac{6}{7}$. Then we see that $\frac{6}{7}=0.857142\ldots$. How do I convert this to base $3$?

Notice that this is $0.\overline{110}$, so it corresponds to $\frac{110_2}{1000_2-1}=\frac 6 7$.
This is how repeating decimals in different bases work: The repeating part can be written as the part that repeats over the difference between the power of the base and $1$.
We need to find where we can do this for $\frac 6 7$ in base $3$. We need to find:
$$\frac 6 7=\frac{a}{3^b-1}$$
We can do this by guessing and checking values of $b$ and seeing if $a$ is an integer. After guessing $b=1,2,3,4,5$, we get non-integer $a$s, but $b=6$ yields $a=624$, so we have:
$$\frac 6 7=\frac{624}{3^6-1}=\frac{212010_3}{1000000_3-1}$$
Thus, the answer is $0.\overline{212010}_3$.

As @BarryCipra, we can solve this without guessing using modular arithmetic:
We want $a=6\frac{3^b-1}{7}$ to be an integer. Leave $6$ out of this since there's no way it's cancelling with the $7$. Thus, $3^b-1$ needs to be a multiple of $7$, so we have: $$3^b-1 \equiv 0 \pmod 7$$ Add $1$ to both sides: $$3^b \equiv 1 \pmod 7$$ Now apply Euler's theorem to get $b=\phi(7)$ as a solution. In this case, $\phi(7)$ is the smallest possible power because $3$ is a primitive root $\pmod 7$.
Note that $\phi(d)$ won't always be the smallest power. For example, in the case of $$\frac 1 8=\frac{a}{3^b-1}$$ If we use $b=\phi(8)=4$, we get $a=10=101_3$ and thus: $$\frac 1 8=0.\overline{0101}_3$$ However, this can be simplified to: $$\frac 1 8=0.\overline{01}_3$$ meaning that $b=2$ was the actual smallest power. However, it still works out because we still find the pattern using $\phi(d)$ and we can simplify the decimal once we find the repeating pattern.
