# Abstract

We explore a generalised logarithm. Normally we have the logarithm which solves for $$x$$ in problems of the form $a=b^x$ giving $x=\mathrm{log}_b(a)$ We can expand this concept by considering the base $$b$$ as a sequence of size $$1$$. We then generalised to any sized set as $a=b_1^x + b_2^x + \cdots b_n^x = \sum_{i=1}^n b_i^x$ to have a solution $\mathrm{log}_{[b_i]}(a)=x$ This then gives us expressions like $\mathrm{log}_{[1,2]}(5)=2\\ \mathrm{log}_{[1,2]}(6)=\frac{\ln 5}{\ln 2}\\ \mathrm{log}_{[1,2]}(7)=\frac{\ln 2 + \ln 3}{\ln 2}\\ \mathrm{log}_{[1,2]}(8)=\frac{\ln 7}{\ln 2}\\ \mathrm{log}_{[1,2]}(9)=3\\ \mathrm{log}_{[1,2]}(10)=\frac{2\ln3}{\ln2}\\ \mathrm{log}_{[1,2]}(17)=4\\$ Which actually corresponds to a general rule $\ln2 \cdot\mathrm{log}_{[1,2]}(n)=\ln(n-1)$

We then not that if a number is repeated we have $x=\mathrm{log}_{[b,b]}(a)=\mathrm{log}_b\left(\frac{a}{2}\right)\\ x=\mathrm{log}_{[b,b,b]}(a)=\mathrm{log}_b\left(\frac{a}{3}\right)\\ x=\mathrm{log}_{[b,b,b,b]}(a)=\mathrm{log}_b\left(\frac{a}{4}\right)\\$

Exploring more size $$2$$ sequences we have $\mathrm{log}_{[1,3]}(10)=2\\ \mathrm{log}_{[1,3]}(11)=\frac{\ln 10}{\ln 3}\\ \mathrm{log}_{[1,3]}(12)=\frac{\ln 11}{\ln 3}\\ \mathrm{log}_{[1,3]}(28)=3\\$

giving generalised rule $\ln3 \cdot \mathrm{log}_{[1,3]}(n)=\ln(n-1)$ and implying generalised rule $\ln b \cdot \mathrm{log}_{[1,b]}(n)=\ln(n-1)$

However, we can then invoke a more complicated $\mathrm{log}_{[2,3]}(2)=0\\ \mathrm{log}_{[2,3]}(5)=1\\ \mathrm{log}_{[2,3]}(13)=2\\ \mathrm{log}_{[2,3]}(14)=2.075994210218454135686379555737786820506\\ \mathrm{log}_{[2,3]}(15)=2.146559601055315555685708181698456455736\\ \mathrm{log}_{[2,3]}(16)=2.212411942612551721235701190647042837254\\ \mathrm{log}_{[2,3]}(17)=2.274134787325801126692290226419467769564\\ \mathrm{log}_{[2,3]}(18)=2.332210076638207392650769064825097217864\\ \mathrm{log}_{[2,3]}(19)=2.387040386805350199779677838034330490689\\ \mathrm{log}_{[2,3]}(20)=2.438965410457542446278518189215003039655\\ \mathrm{log}_{[2,3]}(21)=2.488274376495274934445165074300105158036\\ \mathrm{log}_{[2,3]}(22)=2.535215552035062210571380989640665653591\\ \mathrm{log}_{[2,3]}(23)=2.580003611443195322277567933196601098238\\ \mathrm{log}_{[2,3]}(24)=2.622825421806823602447872527625330439702\\ \mathrm{log}_{[2,3]}(25)=2.663844636013003384060856539904694420856\\ \mathrm{log}_{[2,3]}(26)=2.703205376401337959527922601771322724652\\ \mathrm{log}_{[2,3]}(27)=2.741035216642988538688565217401541061222\\ \mathrm{log}_{[2,3]}(28)=2.777447616245132329235000587523653699195\\ \mathrm{log}_{[2,3]}(29)=2.812543923875739567429215600688647924781\\ \mathrm{log}_{[2,3]}(30)=2.846415037929093629224938682888693196044\\ \mathrm{log}_{[2,3]}(31)=2.879142792311360053767023873678820272534\\ \mathrm{log}_{[2,3]}(32)=2.910801120209828098436925227019535712957\\ \mathrm{log}_{[2,3]}(33)=2.941457037163505572834163597895108057324\\ \mathrm{log}_{[2,3]}(34)=2.971171476057779242124171063934971965086\\ \mathrm{log}_{[2,3]}(35)=3\\$

From a plotting and fitting procedure we find the above data very well fit the relationship $\mathrm{log}_{[2,3]}(x)=\alpha \mathrm{log}(x) + \beta$ with $\alpha \approx 1\\ \beta \approx -0.584343$

this gives $\alpha \ln 13 + \beta = 2\\ \alpha \ln 35 + \beta = 3\\ \alpha = \frac{1}{\ln \frac{35}{13}}=1.00969...\\ \beta = 2-\frac{\ln 13}{\ln \frac{35}{13}}=-0.58981...$

We see that in general$\log_{[b_1\cdots b_n]}(n)=0$ as long as $$b_i>0$$. If one really wanted to retain the property that log any base of $$1$$ is zero, even with the set of number bases, then the generalised log would have to be redefined as the solution to $a=\frac{1}{n}\sum_{i=1}^n b_i^x$

# Infinite Series and Non-integer bases

There remains a great deal of generality. One could consider extreme examples such as $\zeta(1)=\sum_{n=1}^\infty \frac{1}{n} \approx \infty \\ \mathrm{set}[\zeta(1)]=\{\frac{1}{1},\frac{1}{2},\frac{1}{3},\cdots\}$ Then solve some $a=\left(\frac{1}{1}\right)^x + \left(\frac{1}{2}\right)^x+ \left(\frac{1}{3}\right)^x\cdots$ Then we know that $x=\log_{[\zeta(1)]}(a)\\ \log_{[\zeta(1)]}(\zeta(2))=2\\ \log_{[\zeta(1)]}(\zeta(3))=3\\$ Then we see that (at least for the basic support) $\log_{[\zeta(1)]}(\zeta(s))=s\\ \mathrm{inv}[\zeta(s)]=\log_{[\zeta(1)]}(s)$ the log is then the inverse function! We also then have $\log_{[\zeta(b)]}(\zeta(a))=\frac{a}{b}$

We could also use the set $e=\sum_{k=1}^\infty \frac{1}{k!}$ This gives the problem $a=\sum_{k=1}^\infty \frac{1}{(k!)^x}$ and the solution for $$x$$ is $x=\log_{[e]}(a)$ then $\log_{[e]}(\infty)=0\\ \log_{[e]}(e)=1\\ \log_{[e]}(I_0(2))=2\\ \log_{[e]}(_0F_2(;1,1;1))=3\\ \log_{[e]}(2)=\infty$