Discussion

Explanation:

If log_yx = a × log_zy = b × log_xz = ab

⇒ log_yx = ab   ...(1)

⇒ a × log_zy = ab ⇒ log_zy = b   ...(2)

⇒ b × log_xz = ab ⇒ log_xz = a   ...(3)

From (1), (2) and (3), we get
log_yx = log_xz × log_zy

∴  $\frac{\log x}{\log y}$ = $\frac{\log z}{\log x}\times \frac{\log y}{\log z}$ = $\frac{\log y}{\log x}$

∴ (log x)² = (log y)²

∴ log x = ± log y

∴ log x = log y or log x = - log y

∴  x = y or x = $\frac{1}{y}$

∴ ab = log_y x = 1 or -1

Only option (e) does not satisfy this.

Hence, option (e).

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