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For, a^x= b
rnTake log on both the sides
rnlog a^x = log b
rnSince the power can jump before log by the property, log a^2 = 2 log a
rnthe equation becomes,
rnx log a = log b
rntherefore, x = log a / log b
rnIn the same way we can derive, y = log c / log b
rnand z = log a / log c
rnThus performing the multiplication of xyz, we cancel out all the values and we get 1.
Assume a=b=c=1 and x=y=z=2, we get answer as 8. similarly we can get infinite answers, so the answer cannot be definite.
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Take Third Equation that was given: c^z=a
rnSubstitute for the value of \'c\' from the given 2nd equation
rn=> (b^y)^z=a
rn=> b^yz=a ... on simplification
rnagain substituing for the value of \'b\' from first equation
rn=> (a^x)^yz=a
rn=> a^xyz=a ... on simplification
rn=> xyz = 1 ... since bases are equal on both sides, the exponents/powers should be equal
rnWe got our xyz value as 1.