Question

if a=b^2x,b=c^2y and c=a^2z then find the value of xyz

Answer 1

1) a=b^2x

*taking log on both sides
log a= log b^2x = 2x log b

*now taking x on one side
x=log a/2(log b)

similarly
2) y=log b/2(log c)
3) z=log c/2(log a)

so
xyz=(log a/2log b)*(log b/2log c)*( log c/2log a)
on solving
xyz=1/6

Answer 2

Given that,
a=b^2x................................1
b=c^2y................................2
c=a^2z.................................3
We know that,
a^x=N
Then,x=log(base a)N
By above concept we have,
2x=log(base b)a

2x=loga/logb.......................4

2y=log(base c)b

2y=logb/logc......................5

2z=log(base a)c

2z=logc/loga......................6
Multiplying eq4 , eq5 and eq6 we get,
8xyz=(loga/logb)×(logb/logc)×(logc/loga)

8xyz=1

xyz=1/8


Hence value of xyz=1/8