#a^x = b^(1/2), b^y = c^(1/3), c^z = a^(1/2)#. What is the value of #xyz#?

2 Answers
Nov 25, 2016

Answer:

#1/12#

Explanation:

Applying #log# to the equations

#xloga=1/2logb#
#ylogb=1/3logc#
#zlogc=1/2loga#

then

#xyz=1/12( logb logc log a)/(loga logb logc)=1/12#

Nov 26, 2016

#a^x=b^(1/2)#

#=>a^(xy)=(b^y)^(1/2)=(c^(1/3))^(1/2)=c^(1/6)#

#=>(a^(xy))^z=(c^z)^(1/6)=(a^(1/2))^(1/6)=a^(1/12)#

#=>a^(xyz)=a^(1/12)#

#=>xyz=1/12#