Algebra I / Chapter 6: Exponentiation / Logarithms
Proof: Log Product Rule
Let's prove the following theorem:
logb(x ⋅ y) = (logbx) + (logby)
Proof:
Additional Assumptions
1 | logbx = m |
---|---|
2 | logby = n |
3 | logb(x ⋅ y) = o |
# | Claim | Reason |
---|---|---|
1 | bm = x | if logbx = m, then bm = x |
2 | bn = y | if logby = n, then bn = y |
3 | bo = x ⋅ y | if logb(x ⋅ y) = o, then bo = x ⋅ y |
4 | bo = (bm) ⋅ (bn) | if bm = x and bn = y and bo = x ⋅ y, then bo = (bm) ⋅ (bn) |
5 | (bm) ⋅ (bn) = b(m + n) | (bm) ⋅ (bn) = b(m + n) |
6 | bo = b(m + n) | if bo = (bm) ⋅ (bn) and (bm) ⋅ (bn) = b(m + n), then bo = b(m + n) |
7 | o = m + n | if bo = b(m + n), then o = m + n |
8 | logb(x ⋅ y) = (logbx) + (logby) | if logbx = m and logby = n and logb(x ⋅ y) = o and o = m + n, then logb(x ⋅ y) = (logbx) + (logby) |
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