Proof: Log of an Exponential with Different Bases

Let's prove the following theorem:

log_b(x^p) = p ⋅ (log_bx)

Proof:

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Additional Assumptions

1	log_b(x^p) = m
2	log_bx = n

Proof Table

#	Claim	Reason
1	b^m = x^p	if log_b(x^p) = m, then b^m = x^p (Converse of Log Definition)
2	bⁿ = x	if log_bx = n, then bⁿ = x (Converse of Log Definition)
3	b^{(n ⋅ p)} = (bⁿ)^p	b^{(n ⋅ p)} = (bⁿ)^p (power_symmetry_2)
4	(bⁿ)^p = x^p	if bⁿ = x, then (bⁿ)^p = x^p (Substitution)
5	b^{(n ⋅ p)} = b^m	if b^{(n ⋅ p)} = (bⁿ)^p and b^m = x^p and (bⁿ)^p = x^p, then b^{(n ⋅ p)} = b^m (Propagated Transitive Property 3)
6	n ⋅ p = m	if b^{(n ⋅ p)} = b^m, then n ⋅ p = m (Converse of Power Substitution)
7	m = p ⋅ n	if n ⋅ p = m, then m = p ⋅ n (Move Terms Around)
8	log_b(x^p) = p ⋅ n	if log_b(x^p) = m and m = p ⋅ n, then log_b(x^p) = p ⋅ n (Transitive Property of Equality)
9	p ⋅ (log_bx) = p ⋅ n	if log_bx = n, then p ⋅ (log_bx) = p ⋅ n (Substitution)
10	log_b(x^p) = p ⋅ (log_bx)	if log_b(x^p) = p ⋅ n and p ⋅ (log_bx) = p ⋅ n, then log_b(x^p) = p ⋅ (log_bx) (Transitive Property of Equality Variation 1)

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