Let's prove the following theorem:

log_b(x^p) = p ⋅ (log_bx)

Here is an example:

log₂(8³) = 3 ⋅ (log₂8)

log₂(8³) = 3 ⋅ 3

log₂(8³) = 9

Let's check this answer using the definition of logarithms. Since 8³ = 512 ,

log₂(8³) = log₂512

And since 2⁹ = 512

log₂512 = 9

Thus,

log₂(8³) = 9

In both cases, the result is 9.

The key to proving this theorem is the following exponentiation property:

x^m⋅n = (x^m)ⁿ

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 (ConverseOfPowerSubstitution)
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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