math
posted by Shobhit .
If a^x=b^y=c^z and b^2=ac then find value of y.

a^(xyz) = b^(y^2z)
c^(xyz) = b^(xy^2)
(ac)^(xyz) = b^(xy^2+y^2z)
but, ac = b^2, so
(ac)^(xyz) = (b^2)^(xyz) = b^(2xyz)
so,
xy^2 + y^2z = 2xyz
y^2(x+z)  2xyz = 0
y((x+z)y2xz) = 0
so, y=0 or y=2xz/(x+z)
or, from a different angle, since a = b^2/c,
(b^2/c)^x = b^y
xln(b^2/c) = y lnb
y = xln(b^2/c)/lnb
b^y = (b^2/a)^z
y lnb = z ln(b^2/a)
y = z ln(b^2/a)/lnb
pick some values for a,b,c: (2,4,8)
Then x = 3z
y = ln4/(z ln8)
or
y = ln4/(3z ln2)
y = 2xz/(x+z) = 6z^2/4z = 3z/2
so, if z=1, then x=3, y = 3/2
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