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Mathematics 8 Online
OpenStudy (jiteshmeghwal9):

If \(\Large{a^x=b^y=c^z}\) then \(\Large{b^z=ac}\) then \(y=?\)

OpenStudy (anonymous):

ehh... oh jees this isn't that fun. logs ;)

OpenStudy (jiteshmeghwal9):

Are u saying this to me @RONNCC ??

OpenStudy (jiteshmeghwal9):

\[Ans. \frac{2xz}{x+z}\]

OpenStudy (anonymous):

Take a^x=b^y=c^z = k

OpenStudy (jiteshmeghwal9):

ok

OpenStudy (anonymous):

@jiteshmeghwal9 is it b^2 = ac or b^z = ac

OpenStudy (jiteshmeghwal9):

b^z=ac

OpenStudy (anonymous):

b^y = c^z b=c^(z/y) Now, b^z =ac (c^(z/y))^z =ac c^(z^2/y) =ac c^(z^2/y -1)=a c^(z^2/y -1) = c^(z/x) z^2/y -1 = z/x y=(xz^2)/(x+z)

OpenStudy (anonymous):

Yup..thats wat i got

OpenStudy (anonymous):

@jiteshmeghwal9 it shuld be b^2 = ac

OpenStudy (anonymous):

then a , b c are in G.P

OpenStudy (jiteshmeghwal9):

u mean geometric progression

OpenStudy (jiteshmeghwal9):

thanx i gt it :)

OpenStudy (anonymous):

(k^1/y)^2 = k^(z+x)/xz 2/ y = (z+x) / xz y = 2xz / x+z

OpenStudy (jiteshmeghwal9):

thanx both of u @Yahoo! @sauravshakya :)

OpenStudy (anonymous):

Yup..Probably the question is Wrong

OpenStudy (anonymous):

Welcome...:)

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