How do I solve such logarithm I understand that
$\log_b n = x \iff b^x = n$
But all examples I see is with values that I naturally know how to calculate (like $2^x = 8, x=3$)
What if I don't? For example, how do I solve for $x$ when:
$$\log_{1.03} 2 = x\quad ?$$
$$\log_{8} 33 = x\quad ?$$

The logarithm $\log_{b} (x)$ can be computed from the logarithms of $x$ and $b$ with respect to a positive base $k$ using the following formula:
$$\log_{b} (x) = \frac{\log_{k} (x)}{\log_{k} (b)}.$$
So your examples can be solved in the following way with a calculator:
$$x = \log_{1.03} (2) = \frac{\log_{10} (2)}{\log_{10} (1.03)} = 
\frac{0.301}{0.013} = 23.450, $$
$$x = \log_{8} (33) = \frac{\log_{10} (33)}{\log_{10} (8)} = 
\frac{1.519}{0.903} = 1.681.$$
If you know that $b$ and $x$ are both powers of some $k$, then you can evaluate the logarithm without a calculator by the power identity of logarithms, e.g.,
$$x = \log_{81} (27) = \frac{\log_{3} (27)}{\log_{3} (81)} = 
\frac{\log_{3} (3^3)}{\log_{3} (3^4)} = \frac{3 \cdot \log_{3} (3)}{4 \cdot \log_{3} (3)} =
\frac{3}{4}.$$
