# What happens if I select two numbers $a$ and $b$ where $\log a - \log b$ does not give you $\log(a/b)$?

According to the Second Logarithm Law (link via ChiliMath.com), the logarithm of the ratio of two quantities is the logarithm of the numerator minus the logarithm of the denominator. In order words, $$\log a - \log b = \log\frac{a}{b}$$

What happens if I select two numbers $$a$$ and $$b$$ where $$\log a - \log b$$ does not give you $$\log(a/b)$$? What does this mean?

• technically, you can have negative numbers, say a = -1 and b = -1 and let use real (no complex!) numbers only: log(a) and log(b) don't exist, when log(a / b) == 0 Jun 1 '21 at 15:42
• "In order words, log a - log b = log(a/b). What happens if I select two numbers a and b where log a - log b does not give you log(a/b). What does this mean?" Did you find such an example of a and b? If log(a) and log(b) are well defined, i.e., if a > 0 and b > 0, then log(a) - log(b) = log(a/b) is true, and hence you cannot select two numbers a and b such that this is not true.
– Stef
Jun 1 '21 at 15:45
• In your link, they also give these conditions: 1) base > 0 and is not 1; 2) log arguments must be positive and real. If you find a and b such that the Quotient Rule does not hold, then at least one of these conditions is not met. Jun 1 '21 at 15:46
• It means that you did it wrong.
– user65203
Jun 1 '21 at 16:15
• Thank you all for the feedback
– Zizi
Jun 2 '21 at 13:14

Nothing happens, because you cannot select two numbers $$a,b$$ greater than $$0$$ and not get $$\log(\frac{a}{b})$$.
$$\log 0$$ and $$\log x$$ (where $$x$$ is negative) are not defined.