Quote:Original post by SticksandStones
Quote:
m - n = m - n
logbbm - logbbn = logbbm - n (by definition of logarithms, this is equivalent to the above equation)
You have not shown that this is true.
Namely, you have yet to demonstrate that log b^m - log b^n = log b^(m-n)
You're still assuming that which you have to prove.
Well, by definition of logs, the question is b to the what power is bm? and the answer is clearly m because bm = bm. Therefore, logbbm is just m, and the same goes for n and (m - n). Perhaps if I explicitly write out bm = bm it would work?
<edit> Actually, my part might be better rewritten as "since logarithms and exponentiation are inverse operations, logbbm is equivalent to m" </edit>
Quote:
Here's a better way:
log x - log y ? log x/y
Let m = log x <=> b^m = x
Let n = log y <=> b^n = y
m - n ? log x/y
b^(m-n) ? log x/y
log b^(m-n) ? log x/y
log b^m/b^n ? log x/y
log x/y ? log x/y
log x/y = log x/y
Thus: log x - log y = log x/y
I understand your proof, but don't your first and third steps do exactly what you told me I can't do? You change m - n into logbbm-n, which is essentially what I did before.
