Asked by Shobhit
If a^x=b^y=c^z and b^2=ac then find value of y.
Answers
Answered by
Steve
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)y-2xz) = 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
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)y-2xz) = 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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.