Proof: Changing the Log Base

Let's prove the following theorem:

log_bx = (log_kx) / (log_kb)

Proof:

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Additional Assumptions

1	log_kx = i
2	log_kb = j
3	log_bx = k
4	not (j = 0)

Proof Table

#	Claim	Reason
1	kⁱ = x	if log_kx = i, then kⁱ = x (Converse of Log Definition)
2	k^j = b	if log_kb = j, then k^j = b (Converse of Log Definition)
3	b^k = x	if log_bx = k, then b^k = x (Converse of Log Definition)
4	b^k = kⁱ	if b^k = x and kⁱ = x, then b^k = kⁱ (Transitive Property of Equality Variation 1)
5	(k^j)^k = b^k	if k^j = b, then (k^j)^k = b^k (Substitution)
6	(k^j)^k = k^{(j ⋅ k)}	(k^j)^k = k^{(j ⋅ k)} (Power By M Times N)
7	kⁱ = k^{(j ⋅ k)}	if (k^j)^k = b^k and b^k = kⁱ and (k^j)^k = k^{(j ⋅ k)}, then kⁱ = k^{(j ⋅ k)} (Transitive With Four)
8	i = j ⋅ k	if kⁱ = k^{(j ⋅ k)}, then i = j ⋅ k (Converse of Power Substitution)
9	k = i / j	if i = j ⋅ k and not (j = 0), then k = i / j (Divide by Term)
10	log_bx = (log_kx) / (log_kb)	if log_kx = i and log_kb = j and log_bx = k and k = i / j, then log_bx = (log_kx) / (log_kb) (Division Substitution2)

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