Algebra I / Chapter 6: Exponentiation / Logarithms
Proof: Log of an Exponential with Different Bases
Let's prove the following theorem:
logb(xp) = p ⋅ (logbx)
Proof:
Additional Assumptions
1 | logb(xp) = m |
---|---|
2 | logbx = n |
# | Claim | Reason |
---|---|---|
1 | bm = xp | if logb(xp) = m, then bm = xp |
2 | bn = x | if logbx = n, then bn = x |
3 | b(n ⋅ p) = (bn)p | b(n ⋅ p) = (bn)p |
4 | (bn)p = xp | if bn = x, then (bn)p = xp |
5 | b(n ⋅ p) = bm | if b(n ⋅ p) = (bn)p and bm = xp and (bn)p = xp, then b(n ⋅ p) = bm |
6 | n ⋅ p = m | if b(n ⋅ p) = bm, then n ⋅ p = m |
7 | m = p ⋅ n | if n ⋅ p = m, then m = p ⋅ n |
8 | logb(xp) = p ⋅ n | if logb(xp) = m and m = p ⋅ n, then logb(xp) = p ⋅ n |
9 | p ⋅ (logbx) = p ⋅ n | if logbx = n, then p ⋅ (logbx) = p ⋅ n |
10 | logb(xp) = p ⋅ (logbx) | if logb(xp) = p ⋅ n and p ⋅ (logbx) = p ⋅ n, then logb(xp) = p ⋅ (logbx) |
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