MHB Find ab if log_b(a) = log_a(b)

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If log_b(a) = log_a(b) with conditions that a ≠ b, ab > 0, and neither a nor b is 1, then the value of ab is determined to be 1. By letting x = log_a(b) = log_b(a), it follows that a^x = b and b^x = a. Dividing these equations leads to the conclusion that (a/b)^x = (a/b)^{-1}. This implies that ab = 1. The consensus among participants confirms this solution.
My Name is Earl
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I have tried various methods to solve this...

If logb(a) = loga(b) where a != b (!= means does not equal), ab > 0 and neither a nor b are 1, then what is the value of ab?
 
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My Name is Earl said:
I have tried various methods to solve this...

If logb(a) = loga(b) where a != b (!= means does not equal), ab > 0 and neither a nor b are 1, then what is the value of ab?

Let's let:

$$x=\log_a(b)=\log_b(a)$$

Now this implies:

$$a^x=b$$

$$b^x=a$$

Dividing the former by the latter, we obtain:

$$\left(\frac{a}{b}\right)^x=\left(\frac{a}{b}\right)^{-1}$$

What does this imply?
 
MarkFL said:
Let's let:

$$x=\log_a(b)=\log_b(a)$$

Now this implies:

$$a^x=b$$

$$b^x=a$$

Dividing the former by the latter, we obtain:

$$\left(\frac{a}{b}\right)^x=\left(\frac{a}{b}\right)^{-1}$$

What does this imply?

This implies that ab = 1
 
My Name is Earl said:
This implies that ab = 1

Yes, that's what I found as well. (Yes)
 
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