If (log x)/(l + m - 2n) = (log y)/(m + n - 2l) = (log z)/(n + l - 2m), then xyz is equal to :

Question

If \(\frac{\text{log}\,x}{l+m-2n}\) = \(\frac{\text{log}\,y}{m+n-2l}\) = \(\frac{\text{log}\,z}{n+l-2m}\), then xyz is equal to :

(a) 0 

(b) 1 

(c) lmn 

(d) 2

Answer 1

(b) 1

Let \(\frac{\text{log}\,x}{l+m-2n}\) = \(\frac{\text{log}\,y}{m+n-2l}\) = \(\frac{\text{log}\,z}{n+l-2m}\) = k. Then

log x = k(l + m – 2n), log y = k(m + n – 2l); log z = k(n + l – 2m) ⇒ log x + log y + log z = k(l + m – 2n) + k(m + n – 2l) + k(n + l – 2m)

⇒ log(xyz) = 0 ⇒ log(xyz) = log 1 ⇒ xyz = 1.