Inverting the logarithm rule for quotients In school I have been taught that log_a (r/s) =log_ar - log_a(s). Furthermore, it has been said that this rule can be inverted (applied from the right to the left). Then, I came across this: Simplify lg(x−y)−lg(y−x).

Brandon White

Brandon White

Answered question

2022-11-23

Inverting the logarithm rule for quotients
In school I have been taught that log a r s = log a r log a s . Furthermore, it has been said that this rule can be inverted (applied from the right to the left). Then, I came across this:
Simplify lg ( x y ) lg ( y x ).
Note that lg ( x ) means log 10 ( x ). Let's consider x , y R +
The "simplification" yields lg x y y x = lg ( y x y x ) = lg ( 1 )       ( = ln ( 1 ) ln ( 10 ) = π i ln ( 10 ) )
However, if you look at the task, you would get different results depending on whether x > y or y > x. (try it with a calculator or simply google an example...) So, is this rule wrong or do I have to make these "case differentiations" every time I am using this rule?

Answer & Explanation

vihralV5x

vihralV5x

Beginner2022-11-24Added 13 answers

This is one of those cases where there are two seemingly different expressions that express the same value. e i π = 1 and e i π = 1. Both are valid solutions to lg x y y x and which one you arrive at depends on how you break down your expression.

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