3 ms·
Take the log base n of both sides, giving: m = n log_n(m) Because m and n are distinct (ie: m!=n), let's say that m>n. The log term will always be greater th
by anArbitraryOne 3y ago
Take the log base n of both sides, giving:
m = n log_n(m)
Because m and n are distinct (ie: m!=n), let's say that m>n.
The log term will always be greater than one. Imagining the shape of the log function (with base>1), it isn't hard to see that there is one solution.
Not a rigorous proof, but neither was the article.
- phoenixreader 3y agoThis doesn't prove anything, unfortunately. It doesn't even make use of the fact that m,n are integers. If m,n are not restricted to integers, there exists an infinite number of solutions (e.g. m=8.043, n=1.46). The mistake in your proof is that you considered the shape of the log function while implicitly holding m on the left-hand-side constant (i.e. you should have concluded "for a constant m", there is only 1 n satisfying this equality"). However, since m is variable, you have to consider an infinite number of log functions. If I missed something, let me know.
- anArbitraryOne 3y agoYou're right. Not sure what I was thinking. Would be interesting to use a recurrence relation by substituting m