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

1.	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 xa-b . yb-c . zc-a = (a) 0 (b) –1 (c) 1 (d) 2
Answer» (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.

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 xa-b . yb-c . zc-a = (a) 0 (b) –1 (c) 1 (d) 2

Answer»

(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.