Question

if a^x=bc b^y=ca and c^z=ab prove that x/(x+1)+y/(y+1)+z/(z+1)=2 (WITHOUT LOGARITHM)​

Answer 1

Answer:

(a^x)ca = bc(b^y) ---> a^(x+1) = b^(y+1) ---> a^x = b^[x(y+1)/(x+1)]

(b^y)ab = ca(c^z) ---> b^(y+1) = c^(z+1) ---> c^z = b^[z(y+1)/(z+1)]

(c^z)(a^x) = ab.bc = ac.b^2 = b^y.b^2 = b^(y+2)

b^[z(y+1)/(z+1)].b^[x(y+1)/(x+1)] = b^(y+2)

z(y+1)/(z+1) + x(y+1)/(x+1) = y+2

z/(z+1) + x/(x+1) = (y+2)/(y+1)

z/(z+1) + y/(y+1) + x/(x+1) = (y+2)/(y+1) + y/(y+1) = (2y+2)/(y+1) = 2.

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