If (log x)/(a^2+ab+b^2) = (log y)/(b^2+bc+c^2) = (log z)/(c^2+ca+a^2), then x^(a-b). y^(b-c). z^(c-a) =

Question

If \(\frac{log\,x}{a^2+ab+b^2}\) = \(\frac{log\,y}{b^2+bc+c^2}\) = \(\frac{log\,z}{c^2+ca+a^2}\), then x^a-b . y^b-c . z^c-a = 

(a) 0 

(b) –1 

(c) 1 

(d) 2

Answer 1

(c) 1

Let each ratio = k and base = e 

⇒ log_e x = k(a² + ab + b²) 

⇒ (a – b) log_e x = k (a – b) (a² + ab + b²) 

⇒ log_e x^{a – b} = k(a³ – b³) ⇒ x^{a – b} = \(e^{k(a^3-b^3)}\) 

Similarly, y^b-c = \(e^{k(b^3-c^3)}\), z^c-a = \(e^{k(c^3-a^3)}\)

∴ x^a-b . y^b-c . z^c-a = \(e^{k(a^3-b^3)}\). \(e^{k(b^3-c^3)}\) . \(e^{k(c^3-a^3)}\)

= \(e^{k[a^3-b^3+b^3-c^3+c^3-a^3]}\) = e⁰ = 1.