Generalised Multi-base Logarithm

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 \[\log_{[e]}(\infty)=0\\ \log_{[e]}(e)=1\\ \log_{[e]}(I_0(2))=2\\ \log_{[e]}(_0F_2(;1,1;1))=3\\ \log_{[e]}(2)=\infty\]

Other Rules

We have \[9 = 1^2 + \sqrt{8}^2\] then we can write \[\log_{[1,\sqrt{8}]}(9)=2\] However, \[\log_{[1,\sqrt{8}]}(3)=\frac{2}{3}\]

Trying one with non \(1\) base \[5=2^x+3^x\\ 25=2^x+3^x\\ 125=\]

Fermat

Take a notable expression \[a^n + b^n = c^n\] where, \(a,b,c,n\) are all integers. We could express \[n=\log_{[a,b]}(c^n)\] wlog, if \(a\) is \(1\), then we can say \[n=\frac{\ln(c^n -1)}{\ln b}\]