Metamath Proof Explorer


Theorem eflogeq

Description: Solve an equation involving an exponential. (Contributed by Mario Carneiro, 23-Apr-2015)

Ref Expression
Assertion eflogeq ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ B ≠ 0 → e A = B ↔ ∃ n ∈ ℤ A = log ⁡ B + i ⁢ 2 ⁢ π ⁢ n

Proof

Step Hyp Ref Expression
1 efcl ⊢ A ∈ ℂ → e A ∈ ℂ
2 efne0 ⊢ A ∈ ℂ → e A ≠ 0
3 1 2 logcld ⊢ A ∈ ℂ → log ⁡ e A ∈ ℂ
4 efsub ⊢ A ∈ ℂ ∧ log ⁡ e A ∈ ℂ → e A − log ⁡ e A = e A e log ⁡ e A
5 3 4 mpdan ⊢ A ∈ ℂ → e A − log ⁡ e A = e A e log ⁡ e A
6 eflog ⊢ e A ∈ ℂ ∧ e A ≠ 0 → e log ⁡ e A = e A
7 1 2 6 syl2anc ⊢ A ∈ ℂ → e log ⁡ e A = e A
8 7 oveq2d ⊢ A ∈ ℂ → e A e log ⁡ e A = e A e A
9 1 2 dividd ⊢ A ∈ ℂ → e A e A = 1
10 5 8 9 3eqtrd ⊢ A ∈ ℂ → e A − log ⁡ e A = 1
11 subcl ⊢ A ∈ ℂ ∧ log ⁡ e A ∈ ℂ → A − log ⁡ e A ∈ ℂ
12 3 11 mpdan ⊢ A ∈ ℂ → A − log ⁡ e A ∈ ℂ
13 efeq1 ⊢ A − log ⁡ e A ∈ ℂ → e A − log ⁡ e A = 1 ↔ A − log ⁡ e A i ⁢ 2 ⁢ π ∈ ℤ
14 12 13 syl ⊢ A ∈ ℂ → e A − log ⁡ e A = 1 ↔ A − log ⁡ e A i ⁢ 2 ⁢ π ∈ ℤ
15 10 14 mpbid ⊢ A ∈ ℂ → A − log ⁡ e A i ⁢ 2 ⁢ π ∈ ℤ
16 ax-icn ⊢ i ∈ ℂ
17 2cn ⊢ 2 ∈ ℂ
18 picn ⊢ π ∈ ℂ
19 17 18 mulcli ⊢ 2 ⁢ π ∈ ℂ
20 16 19 mulcli ⊢ i ⁢ 2 ⁢ π ∈ ℂ
21 20 a1i ⊢ A ∈ ℂ → i ⁢ 2 ⁢ π ∈ ℂ
22 ine0 ⊢ i ≠ 0
23 2ne0 ⊢ 2 ≠ 0
24 pire ⊢ π ∈ ℝ
25 pipos ⊢ 0 < π
26 24 25 gt0ne0ii ⊢ π ≠ 0
27 17 18 23 26 mulne0i ⊢ 2 ⁢ π ≠ 0
28 16 19 22 27 mulne0i ⊢ i ⁢ 2 ⁢ π ≠ 0
29 28 a1i ⊢ A ∈ ℂ → i ⁢ 2 ⁢ π ≠ 0
30 12 21 29 divcan2d ⊢ A ∈ ℂ → i ⁢ 2 ⁢ π ⁢ A − log ⁡ e A i ⁢ 2 ⁢ π = A − log ⁡ e A
31 30 oveq2d ⊢ A ∈ ℂ → log ⁡ e A + i ⁢ 2 ⁢ π ⁢ A − log ⁡ e A i ⁢ 2 ⁢ π = log ⁡ e A + A - log ⁡ e A
32 pncan3 ⊢ log ⁡ e A ∈ ℂ ∧ A ∈ ℂ → log ⁡ e A + A - log ⁡ e A = A
33 3 32 mpancom ⊢ A ∈ ℂ → log ⁡ e A + A - log ⁡ e A = A
34 31 33 eqtr2d ⊢ A ∈ ℂ → A = log ⁡ e A + i ⁢ 2 ⁢ π ⁢ A − log ⁡ e A i ⁢ 2 ⁢ π
35 oveq2 ⊢ n = A − log ⁡ e A i ⁢ 2 ⁢ π → i ⁢ 2 ⁢ π ⁢ n = i ⁢ 2 ⁢ π ⁢ A − log ⁡ e A i ⁢ 2 ⁢ π
36 35 oveq2d ⊢ n = A − log ⁡ e A i ⁢ 2 ⁢ π → log ⁡ e A + i ⁢ 2 ⁢ π ⁢ n = log ⁡ e A + i ⁢ 2 ⁢ π ⁢ A − log ⁡ e A i ⁢ 2 ⁢ π
37 36 rspceeqv ⊢ A − log ⁡ e A i ⁢ 2 ⁢ π ∈ ℤ ∧ A = log ⁡ e A + i ⁢ 2 ⁢ π ⁢ A − log ⁡ e A i ⁢ 2 ⁢ π → ∃ n ∈ ℤ A = log ⁡ e A + i ⁢ 2 ⁢ π ⁢ n
38 15 34 37 syl2anc ⊢ A ∈ ℂ → ∃ n ∈ ℤ A = log ⁡ e A + i ⁢ 2 ⁢ π ⁢ n
39 38 3ad2ant1 ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ B ≠ 0 → ∃ n ∈ ℤ A = log ⁡ e A + i ⁢ 2 ⁢ π ⁢ n
40 fveq2 ⊢ e A = B → log ⁡ e A = log ⁡ B
41 40 oveq1d ⊢ e A = B → log ⁡ e A + i ⁢ 2 ⁢ π ⁢ n = log ⁡ B + i ⁢ 2 ⁢ π ⁢ n
42 41 eqeq2d ⊢ e A = B → A = log ⁡ e A + i ⁢ 2 ⁢ π ⁢ n ↔ A = log ⁡ B + i ⁢ 2 ⁢ π ⁢ n
43 42 rexbidv ⊢ e A = B → ∃ n ∈ ℤ A = log ⁡ e A + i ⁢ 2 ⁢ π ⁢ n ↔ ∃ n ∈ ℤ A = log ⁡ B + i ⁢ 2 ⁢ π ⁢ n
44 39 43 syl5ibcom ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ B ≠ 0 → e A = B → ∃ n ∈ ℤ A = log ⁡ B + i ⁢ 2 ⁢ π ⁢ n
45 logcl ⊢ B ∈ ℂ ∧ B ≠ 0 → log ⁡ B ∈ ℂ
46 45 3adant1 ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ B ≠ 0 → log ⁡ B ∈ ℂ
47 zcn ⊢ n ∈ ℤ → n ∈ ℂ
48 47 adantl ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ B ≠ 0 ∧ n ∈ ℤ → n ∈ ℂ
49 mulcl ⊢ i ⁢ 2 ⁢ π ∈ ℂ ∧ n ∈ ℂ → i ⁢ 2 ⁢ π ⁢ n ∈ ℂ
50 20 48 49 sylancr ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ B ≠ 0 ∧ n ∈ ℤ → i ⁢ 2 ⁢ π ⁢ n ∈ ℂ
51 efadd ⊢ log ⁡ B ∈ ℂ ∧ i ⁢ 2 ⁢ π ⁢ n ∈ ℂ → e log ⁡ B + i ⁢ 2 ⁢ π ⁢ n = e log ⁡ B ⁢ e i ⁢ 2 ⁢ π ⁢ n
52 46 50 51 syl2an2r ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ B ≠ 0 ∧ n ∈ ℤ → e log ⁡ B + i ⁢ 2 ⁢ π ⁢ n = e log ⁡ B ⁢ e i ⁢ 2 ⁢ π ⁢ n
53 eflog ⊢ B ∈ ℂ ∧ B ≠ 0 → e log ⁡ B = B
54 53 3adant1 ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ B ≠ 0 → e log ⁡ B = B
55 ef2kpi ⊢ n ∈ ℤ → e i ⁢ 2 ⁢ π ⁢ n = 1
56 54 55 oveqan12d ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ B ≠ 0 ∧ n ∈ ℤ → e log ⁡ B ⁢ e i ⁢ 2 ⁢ π ⁢ n = B ⋅ 1
57 simpl2 ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ B ≠ 0 ∧ n ∈ ℤ → B ∈ ℂ
58 57 mulridd ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ B ≠ 0 ∧ n ∈ ℤ → B ⋅ 1 = B
59 52 56 58 3eqtrd ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ B ≠ 0 ∧ n ∈ ℤ → e log ⁡ B + i ⁢ 2 ⁢ π ⁢ n = B
60 fveqeq2 ⊢ A = log ⁡ B + i ⁢ 2 ⁢ π ⁢ n → e A = B ↔ e log ⁡ B + i ⁢ 2 ⁢ π ⁢ n = B
61 59 60 syl5ibrcom ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ B ≠ 0 ∧ n ∈ ℤ → A = log ⁡ B + i ⁢ 2 ⁢ π ⁢ n → e A = B
62 61 rexlimdva ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ B ≠ 0 → ∃ n ∈ ℤ A = log ⁡ B + i ⁢ 2 ⁢ π ⁢ n → e A = B
63 44 62 impbid ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ B ≠ 0 → e A = B ↔ ∃ n ∈ ℤ A = log ⁡ B + i ⁢ 2 ⁢ π ⁢ n