Why does n*log(-a) != log([-a]^n).
For example, suppose n = 2, and a =4.

Question

Why does n*log(-a) != log([-a]^n).
For example, suppose n = 2, and a =4.

anonymous · Answer

$$n \log (-a) \neq \log([-a]^n)$$

e.g. n =2, a = 4
$$n \log (-a) \neq \log([-a]^n)$$
$$2 \log (-4) \neq \log([-4]^2)$$
$$2 \log (-4) \neq \log(16)$$
LHS is undefined and right hand side has a value.
Why do the log laws not seem to work in this case?

empty · Answer

The reason is you're assuming $a$ is a negative number and then plugging in a positive number. It will work if you plug in negative values for a.

empty · Answer

Ok ok, maybe that's kind of a cheap answer, so I'll say it another way:

$$\log(4)=\log((-2)^2)\ne2\log(-2)$$

So unfortunately negative numbers in logarithms are mostly nonsense, but you can find answers for what they mean if you are working with complex numbers. But at that point, then you will find that the logarithm is really multi-valued!

zzr0ck3r · Answer

i agree. you put in positive on the rhs, but the lhs is undefined in the example

It is sort of like starting with 2=3 and then concluding that 1=0 and then questioning addition

anonymous · Answer

Thanks guys, so the law just basically assumes that a and n are positive.