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Explanation:

If logyx = a × logzy = b × logxz = ab

⇒ logyx = ab   ...(1)

⇒ a × logzy = ab ⇒ logzy = b   ...(2)

⇒ b × logxz = ab ⇒ logxz = a   ...(3)

From (1), (2) and (3), we get
logyx = logxz × logzy

∴  log xlog y = log zlog x×log ylog z = log ylog x

∴ (log x)2 = (log y)2

∴ log x = ± log y

∴ log x = log y or log x = - log y

∴  x = y or x = 1y

∴ ab = logy x = 1 or -1

Only option (e) does not satisfy this.

Hence, option (e).

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