What is the fractal dimension/Hausdorff dimension of a Koch's snowflake? I have  found that the fractal dimension of a self-similar object is:
$$\text{fractal dimension} = \frac{\log(\text{number of self-similar pieces})}{\log(\text{magnification factor})} $$
See here details for the formula from above.
Therefore, using that formula can we conclude that the dimension of a Koch snowflake is $\frac{\log 6 }{\log 3}$?
When I searched online it's dimension appeared to be $\frac{\log 4}{\log 3}$. Why? Where do things fall apart with the first formula?
 A: First off, the solid Koch Snowflake is, in fact, self-similar; it consists of seven copies of itself - six of which, shown in gray in figure below, are scaled by the factor $1/3$ and one of which, shown in red in the figure below, is scaled by the factor $1/\sqrt{3}$.

The formula that you mention,
$$
\text{dimension} = \frac{\log(\text{number of self-similar pieces})}{\log(\text{magnification factor})},
$$
works only for simpler sets, where all the pieces have the same scaling factor. More generally, a self-similar set can consist of $N$ copies of itself scaled by the factors $\{r_1,r_2,\ldots,r_N\}$. In this case, the similarity dimension of the set is defined to be the unique value of $s>0$ such that
$$
 r_1^s + r_2^s + \cdots + r_N^s = 1.
$$
Note that if $r_1=r_2=\cdots=r_N = r$, then the equation simplifies to $N r^s=1$. In this case, you can solve for $s$ to get the simpler formula.
For the solid Koch curve, we expect the dimension to be 2. In fact, if we set $s=2$ and use the scaling factors for the solid Koch flake, we get.
$$
  6 (1/3)^2 + (1/\sqrt{3})^2 = 2/3+1/3 = 1,
$$
as expected.
