apply le_of_eq_of_le
case a
b n : ℕ
hb : 1 < b
hn : 1 ≤ n
⊢ log b (b * n) ≤ log b n + 1

case a
b n : ℕ
hb : 1 < b
hn : 1 ≤ n
⊢ log b n + 1 ≤ log b (b * n)
case a.h₁
b n : ℕ
hb : 1 < b
hn : 1 ≤ n
⊢ log b (b * n) = ?a.b✝

case a.h₂
b n : ℕ
hb : 1 < b
hn : 1 ≤ n
⊢ ?a.b✝ ≤ log b n + 1

case a.b
b n : ℕ
hb : 1 < b
hn : 1 ≤ n
⊢ ℕ

case a
b n : ℕ
hb : 1 < b
hn : 1 ≤ n
⊢ log b n + 1 ≤ log b (b * n)