Question

If a^1/x=b/1y=c^1/a,b^2=ac, then find the value of x+y/2y​

Answer 1

Answer:

If A^x=b^y=c^z and b^2=AC, then what is the value of y?

ax=by=>a=byx

by=cz=>c=byz

b2=ac=byx∗byz=byx+yz

Since the base are same, powers must be equal.

=>2=yx+yz=y∗(1x+1z)=y(x+z)xz

=>y=2xzx+z

A^x= b^y = c^z , from this if we like to express A & c in terms of b —-

A^x = b^y, => A = b^(y/x) , similarly , as c^z = b^y Therefore , c = b^(y/z). It is also given that ——

b^2 = Ac,

or,b^2 = {b^(y/x)}*{b^(y/z)} ( putting the value of A & c defined in terms of b )

Or, b^2= b^(y/x+y/z)

or, b^2 = b^{( yz +xy)/xz }

or, (xy + yz)/xz =2 (as base on the both side equals (b) , power is equal)

or, xy + yz = 2xz

or, y(x+z) = 2xz

So, y = 2xz/(x+z)

Answer 2

Step-by-step explanation:

If A^x=b^y=c^z and b^2=AC, then what is the value of y?

ax=by=>a=byx

by=cz=>c=byz

b2=ac=byx∗byz=byx+yz

Since the base are same, powers must be equal.

=>2=yx+yz=y∗(1x+1z)=y(x+z)xz

=>y=2xzx+z

A^x= b^y = c^z , from this if we like to express A & c in terms of b —-

A^x = b^y, => A = b^(y/x) , similarly , as c^z = b^y Therefore , c = b^(y/z). It is also given that ——

b^2 = Ac,

or,b^2 = {b^(y/x)}*{b^(y/z)} ( putting the value of A & c defined in terms of b )

Or, b^2= b^(y/x+y/z)

or, b^2 = b^{( yz +xy)/xz }

or, (xy + yz)/xz =2 (as base on the both side equals (b) , power is equal)

or, xy + yz = 2xz

or, y(x+z) = 2xz

So, y = 2xz/(x+z)

HOPE IT HELPS YOU OUT ✌

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