OK, why not like this: 1/3 / √ 32 = 2^n
Hello guest from answer #1!
\(\color{BrickRed}\frac{1}{3\cdot \sqrt{32}}=2^n\\ 3\cdot \sqrt{32}=2^{-n}\\ 3\cdot 2^{2.5}=2^{-n}\\ 3=\frac{2^{-n}}{2^{2.5}}\\ 3=2^{-n-2.5}\)
\(\frac{ln3}{ln2}=-n-2.5\)
\(n=-2.5-\frac{ln3}{ln2}\)
\(n=-4.08496\)
!