Proof: Log Product Rule

Let's prove the following theorem:

This theorem allows us to convert a log of a product to a sum of the log of each operand.

For example:

log₂(8 ⋅ 16) = (log₂8) + (log₂16)

log₂(8 ⋅ 16) = 3 + 4

log₂(8 ⋅ 16) = 7

Before calculators were invented, multiplying large numbers would take a long time. People used the Log Product Rule to convert multiplication problems to addition problems which are easier to solve.

Let's try multiplying two numbers using the following algorithm:

1. Calculate log base 10 of both operands using a log table.
2. Add the results.
3. Calculate 10 to the power of the result from step 2 using an "anti-log" table.

For example, suppose we want to calculate 14⋅18.

First, we calculate:

log₁₀(14 ⋅ 18)

This is equal to

(log₁₀14) + (log₁₀18)

Then, the log table tells us that:

log₁₀14 = 1.146

log₁₀18 = 1.255

Thus:

(log₁₀14) + (log₁₀18) = 2.401

Finally, the anti-log table tells us that:

10^2.401 = 251.77

14⋅18 = 252, so we came quite close to the answer.

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

x^m ⋅ xⁿ = x^(m+n)

Proof:

Proof Table

#	Claim	Reason
1	bm = x	if logbx = m, then bm = x (Converse of Log Definition)
2	bn = y	if logby = n, then bn = y (Converse of Log Definition)
3	bo = x ⋅ y	if logb(x ⋅ y) = o, then bo = x ⋅ y (Converse of Log Definition)
4	bo = (bm) ⋅ (bn)	if bm = x and bn = y and bo = x ⋅ y, then bo = (bm) ⋅ (bn) (Multiplication Substitution)
5	(bm) ⋅ (bn) = b(m + n)	(bm) ⋅ (bn) = b(m + n) (power_symmetry)
6	bo = b(m + n)	if bo = (bm) ⋅ (bn) and (bm) ⋅ (bn) = b(m + n), then bo = b(m + n) (Transitive Property of Equality)
7	o = m + n	if bo = b(m + n), then o = m + n (ConverseOfPowerSubstitution)
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) (Addition Substitution)

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