Proof: Changing the Log Base

Let's prove the following theorem:

log_bx = (log_cx) / (log_cb)

We can use this rule to change the base of the logarithm. Before calculators were invented, people used log tables to calculate the log of a value. But the log tables only contained answers for certain bases (usually 10). Thus, people used this rule to convert the log operation to the right base. Even now, many calculators only know how to compute logs in base 10 and e (natural log), so we still have to convert log operations to one of these two bases.

For example, suppose that we want to calculate log₂8 .

Our calculator does not know how to compute base 2, so we convert the expression as follows:

log₂8 = (log₁₀8) / (log₁₀2)

We can now use the calculator to evaluate log₁₀8 and log₁₀2. Or we can simplify the right side by substituting 2³ for 8 and using the "Log of an Exponential" rule. Then:

log₂8 = (log₁₀(2³)) / (log₁₀2)

log₂8 = (3 ⋅ (log₁₀2)) / (log₁₀2)

log₂8 = 3

We can also go the other way. Suppose that we are trying to compute

(log₁₀49) / (log₁₀7)

Suppose that we don't have a calculator, so we cannot evaluate either log₁₀49 or log₁₀7 , but we can convert this division problem to:

log₇49

And we know that log₇49 is 2. Therefore,

(log₁₀49) / (log₁₀7) = 2

High-Level Proof

Suppose that:

log_cx = i
log_cb = j
log_bx = k

Then:

cⁱ = x
c^j = b
b^k = x

Thus:

cⁱ = b^k

Also:

(c^j)^k = b^k

Which means:

cⁱ = c^{j ⋅ k}

Thus:

i = j ⋅ k

Which we can transform to:

k = i / j

Proof:

View as a tree | View dependent proofs | Try proving it

Assumptions

1	log_cx = i
2	log_cb = j
3	log_bx = k
4	not (j = 0)

Proof Table

#	Claim	Reason
1	cⁱ = x	if log_cx = i, then cⁱ = x (Converse of Log Definition)
2	c^j = b	if log_cb = j, then c^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 = cⁱ	if b^k = x and cⁱ = x, then b^k = cⁱ (Transitive Property of Equality Variation 1)
5	(c^j)^k = b^k	if c^j = b, then (c^j)^k = b^k (Substitution)
6	(c^j)^k = c^{(j ⋅ k)}	(c^j)^k = c^{(j ⋅ k)} (Power of a Power)
7	cⁱ = c^{(j ⋅ k)}	if (c^j)^k = b^k and b^k = cⁱ and (c^j)^k = c^{(j ⋅ k)}, then cⁱ = c^{(j ⋅ k)} (Transitive With Four)
8	i = j ⋅ k	if cⁱ = c^{(j ⋅ k)}, then i = j ⋅ k (ConverseOfPowerSubstitution)
9	k = i / j	if i = j ⋅ k and not (j = 0), then k = i / j (Divide by Term)
10	log_bx = (log_cx) / (log_cb)	if log_cx = i and log_cb = j and log_bx = k and k = i / j, then log_bx = (log_cx) / (log_cb) (Division Substitution2)

Previous Lesson Next Lesson

Comments

Please log in to add comments