Proof: Changing the Log Base
Let's prove the following theorem:
logbx = (logkx) / (logkb)
Proof:
Additional Assumptions
1 | logkx = i |
---|---|
2 | logkb = j |
3 | logbx = k |
4 | not (j = 0) |
# | Claim | Reason |
---|---|---|
1 | ki = x | if logkx = i, then ki = x |
2 | kj = b | if logkb = j, then kj = b |
3 | bk = x | if logbx = k, then bk = x |
4 | bk = ki | if bk = x and ki = x, then bk = ki |
5 | (kj)k = bk | if kj = b, then (kj)k = bk |
6 | (kj)k = k(j ⋅ k) | (kj)k = k(j ⋅ k) |
7 | ki = k(j ⋅ k) | if (kj)k = bk and bk = ki and (kj)k = k(j ⋅ k), then ki = k(j ⋅ k) |
8 | i = j ⋅ k | if ki = k(j ⋅ k), then i = j ⋅ k |
9 | k = i / j | if i = j ⋅ k and not (j = 0), then k = i / j |
10 | logbx = (logkx) / (logkb) | if logkx = i and logkb = j and logbx = k and k = i / j, then logbx = (logkx) / (logkb) |
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