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:
$$\log_{1.03} 2 = x\quad ?$$
$$\log_{8} 33 = x\quad ?$$
 A: 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.,
$$x = \log_{81} (27) = \frac{\log_{3} (27)}{\log_{3} (81)} = 
\frac{\log_{3} (3^3)}{\log_{3} (3^4)} = \frac{3 \cdot \log_{3} (3)}{4 \cdot \log_{3} (3)} =
\frac{3}{4}.$$
A: method 1: use a calculator
method 2 (more fun): $\log_b a=\frac{\ln b}{\ln a}$
To calculate natural logs, if $|x|<1$ use the power series $\ln (x+1)=x-\frac{x^2}{2}+\frac{x^3}{3}-...$ and if not find the log of the reciprocal and subtract from zero. Powers of $x$ can be calculated by convolving as power series in $10$. 
A: There's actually nothing to solve here. You have an exact expression for $x$ in each case- you may not know exactly what number it corresponds to, just like you might not know what the square root of three equals when you solve something like $x^2=3$ but it's still just a number- in the first case, it would be the power to which I raise 1.03 to get 2 which is approximately 23.45
A: $$\log_bn=\frac{\ln n}{\ln b}=\frac{\log_{10}n}{\log_{10}b}=\frac{\log_2n}{\log_2b}=\ldots$$
Edit: Now added in a comment:

I want to find the value without a calculator.

...Which is an entirely new take on the question.
To get $x=\log_{1.03}2$, one could compute the successive powers of $1.03$. If $1.03^n\lt2\lt1.03^{n+1}$, then $n\lt x\lt n+1$. The rest depends on your ability to compute $1.03^n$ for $n$ in the $20$-$30$ range... but this yields $n=23$.
Another approach is to use power series, in this case $x=y/z$ with $y=\ln2$ and $z=\ln1.03$. Using the expansion $\log1+t=\sum\limits_{n\geqslant1}(-1)^{n-1}\frac{t^n}n$, one gets
$$
y=-\ln\left(1-\tfrac12\right)=\sum_{n\geqslant1}\frac1{n2^n},\qquad
z=\ln(1+.03)=\sum\limits_{n\geqslant1}(-1)^{n-1}\frac{(.03)^n}n.
$$
Keeping $9$ terms in $y$ and $2$ terms in $z$ yields an error of at most $10^{-4}$.
A: Starting with :$$\log_{1.03} 2 = x$$This is exactly the same as $$1.03^x=2$$ now take logarithms of both sides$$x \times \log(1.03)=\log(2)$$ Now, use your calculator
