(k1/6)^-2/(k^5)1/3=

Question

(k 1/6) ^-2  over

(k^5) 1/3   =                   Assume all variables are positive.  

Answer 1

the problem looks a little confusing the way it is written

for one thing, is the exponent 2 or -2 ?

 

however, (k 1/6)^-2=1/(k 1/6)^2=1/[(k^2)/36)]=36/k^2

(k^5)1/3=k^5/3

[36/(k^2)]/[(k^5)/3]=[36/(k^2)]*[3/(k^5)]=108/(k^7)

 

if the exponent is 2 we get (k^2/36)*(3/k^5)=1/12k^3

 

Answer 2

Hi David,

 

(k 1/6)^-2 is the same as 1/(k 1/6)^2 which =1/k^2/36 which is 36/k^2 and (k^5)1/3 is just (k^5)/3

 

36/k^2 over k^5/3 can be rewritten as a multiplication problem by inverting the 2nd fraction

 

so we end up with 36/k^2*3/k^5=108/k^7

 

This is different than what I wrote at 130 am because of a mistake I made in the 1st term.  When I saw Arthur's answer I found my mistake and edited this one so it didn't confuse you.

 

Hope this helps