How do you evaluate #\frac{(2p^{3}q^{4})^{4}}{(-3q^{5})^{2}}\div \frac{(4p^{2}q)^{2}}{9}=\frac{p^{a+b}}{q^{a-b}}#?

Question

965 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-15T14:20:07

#b=6#
#a=2#

#(p^(a+b))/(q^(a-b))=(p^(2+6))/(q^(2-6))=(p^(8))/(q^(-4))#

#((2p^3q^4)^4)/((-3q^5)^2)-:((4p^2q)^2)/9=(p^(a+b))/(q^(a-b))#

#(2^4*p^((3*4))*q^((4*4)) )/((-3)^2*q^((5*2)))*9/(4^2*p^((2*2))*q^2)#

#(cancel16*p^(12)*q^(16) )/(cancel9*q^(10))*cancel9/(cancel16*p^(4)*q^2)#

#(p^(12)*q^(16) )/(q^(10))*1/(p^(4)*q^2)#

#(p^(12)*q^(16) )/(q^(10)*p^(4)*q^2)#

#(p^(12)*q^(16) )/(q^((10+2))*p^(4))#

#(p^((12-4)))/(q^((12-16)))=(p^(8))/(q^(-4))#

#a+b=8#
#a-b=-4quad=>quada=b-4#

#b-4+b=8quad=>quad2b=12quad=>quadb=6#
#a=b-4quad=>quada=6-4quad=>quada=2#