The Wikipedia page on the natural logarithm says: 'Logarithms can be defined to any positive base other than 1, not only e. However, logarithms in other bases differ only by a constant multiplier from the natural logarithm, and are usually defined in terms of the latter.' This seems to suggest that only natural logarithms have this property, namely: $$\tag{0}log_a(x)=c.log_e(x)$$ where $x$ can vary and yet $c$ remains constant. This question follows on from this one about the result: $$\tag{1}\int\frac{c}{x}=c.log_e(x) + C$$ given by WolframAlpha, where they explicitly state that it is the natural logarithm being used. However, my reading of the Wikipedia proof of the differentiation of $ln(x)$ (which is actually base agnostic) led me to believe that the above equation can be written: $$\tag{2}\int\frac{log_a(e)}{x}=log_a(x)+C$$ If $a=e$ they do of course match; non-trivially the implication, from $(0)$, is that: $$c=log_a(e)$$ $$\frac{log_a(x)}{log_a(e)}=log_e(x)$$ (See Wikipedia, Logarithm, Change of base). Algebraically, this all checks out; however we began by assuming the property that connected the two integrals, so we must ask the question - does that property really only apply to natural logarithms? And, could the Wolfram result be rewritten without specifying a base at all?
Consider the following test:
from math import log #log is ln in python unless a 2nd argument gives a base
print()
print('log(18,1.5)', log(18,1.5))
print('log(18,3)', log(18,3))
print('log(18,1.5)/log(18,3)', log(18,1.5)/log(18,3))
print()
print('log(14,1.5)', log(14,1.5))
print('log(14,3)', log(14,3))
print('log(14,1.5)/log(14,3)', log(14,1.5)/log(14,3))
print()
which produces the results:
7.12853 6.50872
2.63093 2.40217
2.70951 2.70951
The third and sixth results agreed exactly (i.e. in every decimal place) although one had one more place than the other. What do these results show? We are trying to test the theory that: $$log_a(x)=c.log_b(x)$$ where $x$ can vary for bases (b) other than $e$. So: $$\frac{log_a(x_1)}{log_b(x_1)}=\frac{log_a(x_2)}{log_b(x_2)}$$ because they both equal $c$. This is exactly what was being tested. Rearranging: $$\frac{log_a(x_1)}{log_a(x_2)}=\frac{log_b(x_1)}{log_b(x_2)}$$ $$log_{x2}(x_1)=log_{x2}(x_1)$$ And so $1=1$! The last step invokes the aforementioned change of base rule (which can be proven from $x=b^{log_b(x)}$ and the power law of logarithms). The upshot of all this is that it is very misleading to say that $\int c/x = c.log_e(x)+C$. It could use any logarithm at all.
EDIT: to complete the differentiation of $log(x)$ take $(2)$ above (that is how it 'comes out'), convert it to: $$\int\frac{log_a(e)}{x}=log_a(e).log_e(a).log_a(x)+C$$ from David K (2), then do $log_e(a).log_a(x)=log_e(x)$ and $log_a(e)=c$ (because $a$ is no longer related to $x$) to obtain $(1)$ above. Only the natural logarithm allows us to do this - to find the integral of $c/x$ we find $c.ln(x)$ instead of having to work out the base of $log_a(e)=c$ and then the value of $log_a(x)$. Before computers this would probably have been the deciding factor in choosing the best technique (because natural logarithms were tabulated).