Why are $e^{-4\ln(-x)+4}$ and $e^{-4\ln(-x)}$ positive?

Question

Algebraically, why is $e^{-4\ln(-x)+4}$ a positive result but $e^{-4\ln(-x)}$ a positive result?

It seems like with logarithmic properties, they should both be negative, but Wolfram Alpha shows the graph and states that it must be positive. Is there a logarithmic identity to explain this discrepancy? Thanks much!

Answer 1

$e^y > 0$ for all $y$ which means that it is also the case for $y=-4ln(-x)+4$ and $y=-4ln(-x)$ means $e^{-4ln(-x)+4} > 0$ and $e^{-4ln(-x)>0} >0$ there where they are defined.

Answer 2

You can write the former as $$e^{-4\ln(-x) + 4} = e^{-4\ln(-x)} \cdot e^4 = e^{\ln((-x)^{-4})} \cdot e^4 = \frac{e^4}{x^4}.$$ Notice that this will always be nonnegative.

Similarly, you can write $$e^{-4\ln(-x)} = e^{\ln((-x)^{-4})} = \frac{1}{x^4},$$ which will still be nonnegative.