If log_b 6 = .4040 then b^-.4040 is equal to...I am not to sure about how to solve this.
First, you should start a new thread for a new question.
Note that $\displaystyle \log(\sqrt{a})=\tfrac{1}{2}\log(a)$.
$\displaystyle \log(20)=\log(10)+\log(2)$
Now contrary to modern trends it seems that here $\displaystyle \log$ means $\displaystyle \log_{10}$ so $\displaystyle \log(10)=1$.
No, I am absolutely not angry. I inferred that impression on my first reading of your post; but, after re-reading it, I have found that impression to be wrong. I am terrible sorry. I sensitive when it comes to scorning my mathematical knowledge; and I know it is scarce, due to myself having studied it a little later in life than normal, and that is why I was a bit jumpy. Again, sorry.