Taylor series expansion of base 2 logarithms Sorry for the noob question, but I've been hitting my head against the wall on this for a while.
I am looking for a Taylor series expansion of a logarithm other than the natural logarithm $ln(x)$. It seems that every piece of literature I've been going through treats solely the natural logarithm and not logarithms in other bases. 
In particular, I would like to know the Taylor series corresponding to the binary logarithm $log_2(x)$. For instance, how would I go about calculating $log_2(3)$ using the Taylor series?
Thanks!
 A: Find the Taylor series of $\log_2(x)=\frac{\ln(x)}{\ln(2)}$ at a point close to $x=3$, for instance $x=\frac{5}{2}$ 
$$ \log_2(x)= \frac{1}{\ln(2)}\left(\ln  \left( 5 \right) -\ln  \left( 2 \right) +{\frac {2}{5}} \left( x
-{\frac {5}{2}} \right) -{\frac {2}{25}} \left( x-{\frac {5}{2}}
 \right) ^{2}+{\frac {8}{375}} \left( x-{\frac {5}{2}} \right) ^{3}+O
 \left(  \left( x-{\frac {5}{2}} \right) ^{4} \right)\right).$$
Now, subs $x=3$ in the above approximation series gives

$$ \log_2(3)\sim 1.585460388  $$

with absolute error

$$ 0.000497887. $$

Note: You can find the Taylor series at the point $x=e$ which makes the derivation easier

$$ \log_2(x) \sim \frac{1}{\ln(2)} \left( 1+\frac{1}{e}(x-e)-\frac{1}{2e^2}(x-e)^2 + \frac{1}{3e^3}(x-e)^3\right). $$

A: $\log_2(x)$=$\ln(x)/\ln(2)$. Then expand the $\ln(x)$.
A: Base conversion of $\ln(\cdot)$ is correct analytically, but, curiously, in error in binary computers. To see the actual recursively convergent expansion of the binary logarithm (the binary "equivalent" of the Taylor series), a good starting point is the Wiki article binary logarithm, and following its links and associated pages.
