+0  
 
0
1089
3
avatar

Help?

 Oct 15, 2020
 #1
avatar
0

By the way, it should actually be:

work ou the value of n if:

1/3 / √ 32 = 2^n

 Oct 15, 2020
 #2
avatar+14903 
+1

1/3 √ 32 = 2^n

 

Hello Guest!

 

\(\frac{1}{3}\cdot \sqrt{2^5}=2^n\\ \sqrt{2^5}=3\cdot 2^n\\ 2^{\frac{5}{2}-n}=3\)

\(ln2\cdot ({\frac{5}{2}-n})=ln3\)

\(\frac{5}{2}-n=\frac{ln3}{ln2}\)

\(n=\frac{5}{2}-\frac{ln3}{ln2}\)

\(n=0.15037\)

laugh  !

 Oct 15, 2020
edited by asinus  Oct 15, 2020
 #3
avatar+14903 
+1

OK, why not like this:  1/3 / √ 32 = 2^n

 

Hello guest from answer  #1!

 

\(\color{BrickRed}\frac{1}{3\cdot \sqrt{32}}=2^n\\ 3\cdot \sqrt{32}=2^{-n}\\ 3\cdot 2^{2.5}=2^{-n}\\ 3=\frac{2^{-n}}{2^{2.5}}\\ 3=2^{-n-2.5}\)

\(\frac{ln3}{ln2}=-n-2.5\)

\(n=-2.5-\frac{ln3}{ln2}\)

\(n=-4.08496\)

laugh  !

 Oct 15, 2020
edited by asinus  Oct 15, 2020

3 Online Users

avatar
avatar
avatar