đŸ Archived View for gemini.locrian.zone âș misc âș powers_of.gmi captured on 2023-11-14 at 08:10:55. Gemini links have been rewritten to link to archived content
-=-=-=-=-=-=-
Original post on r/askmath by u/Arroway97
Recently I was thinking about how you can take any number k where k is any real number not equal to â1, 0, or 1 and find some power x such that kx = n, where x is any real number and n is any real number thatâs always positive when k is positive, negative when k is negative and x is odd, and positive when k is negative and x is even or 0 (the usual power rules).
If you try to extend this to k = 1, this doesnât hold of course, but if you start with k = 1.1 and continue getting infinitely closer to k = 1, x starts to approach an infinitely long number. For k = 1.1 and n = 2, i.e. 1.1^x = 2, the value of x is 7.272540897... But for 1.01^x = 2, x = 69.6607168... For 1.001^x = 2, x = 693.493696... For 1.0001^x = 2, x = 6931.81837... For 1.000000001^x = 2, x = 693147123... And so on. I know that in order to express this it would use a limit but I havenât been in calculus for a long while so I canât figure out how to actually formulate it. This can also be extended to other values of n besides 2 for all other reals, suggesting that there is a unique, infinitely long number that solves the equation for each real number value of n. You can even approach 1 from the left hand side and you would find negative solutions for x that mirror the positive solutions until you reach 1 where the equation becomes undefined.
How would you formally express the number that satisfies the equation for some number n. How do these solutions map to the set of reals? Have these numbers been studied before? How do you even calculate an infinitely long number that grows in size on both sides of the decimal point? Would the resulting solution as you get asymptomatically close to 1 be a whole number or one with infinite numbers on either side of the decimal point? I have so many questions
You can grasp you "infinitely long"-number if you divide it by powers of 10. You might see that the solution of 1.000001x = 2 is x=691347.52... If you divide x by 100000 you get 0.691347... which seems very close to ln(2)! In fact: The solution to (1+0.1n)^x = 2 is always ln(2)/ln(10-n +1). This is just ln(2) multplied with approximately some power of 10. If you divide this solution by the corresponding power of 10 you can form a proper limit and this limit is exactly ln(2). So in some sense your sequence has the limit ln(2) but with the decimal point âmoved by an infinite amount to the rightâ.