English
For base b > 1 and n > 0, log_b n = log_b(n+1) iff b^{log_b(n+1)} ≠ n+1.
Русский
Для основания b > 1 и n > 0 верно: log_b n = log_b(n+1) тогда и только тогда, когда b^{log_b(n+1)} ≠ n+1.
LaTeX
$$$ (1 < b) \land (n \neq 0) \Rightarrow \log_b n = \log_b(n+1) \iff b^{\log_b(n+1)} \neq n+1 $$$
Lean4
theorem log_eq_log_succ_iff {b n : ℕ} (hb : 1 < b) (hn : n ≠ 0) : log b n = log b (n + 1) ↔ b ^ log b (n + 1) ≠ n + 1 :=
by
rw [ne_eq, ← log_lt_log_succ_iff hb hn, not_lt]
simp only [le_antisymm_iff, and_iff_right_iff_imp]
exact fun _ ↦ log_monotone (le_add_right n 1)