Metamath Proof Explorer


Theorem eflogeq

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

Ref Expression
Assertion eflogeq A B B 0 e A = B n A = log B + i 2 π n

Proof

Step Hyp Ref Expression
1 efcl A e A
2 efne0 A e A 0
3 1 2 logcld A log e A
4 efsub A log e A e A log e A = e A e log e A
5 3 4 mpdan A e A log e A = e A e log e A
6 eflog e A e A 0 e log e A = e A
7 1 2 6 syl2anc A e log e A = e A
8 7 oveq2d A e A e log e A = e A e A
9 1 2 dividd A e A e A = 1
10 5 8 9 3eqtrd A e A log e A = 1
11 subcl A log e A A log e A
12 3 11 mpdan A A log e A
13 efeq1 A log e A e A log e A = 1 A log e A i 2 π
14 12 13 syl A e A log e A = 1 A log e A i 2 π
15 10 14 mpbid A A log e A i 2 π
16 ax-icn i
17 2cn 2
18 picn π
19 17 18 mulcli 2 π
20 16 19 mulcli i 2 π
21 20 a1i A i 2 π
22 ine0 i 0
23 2ne0 2 0
24 pire π
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 i 2 π 0
30 12 21 29 divcan2d A i 2 π A log e A i 2 π = A log e A
31 30 oveq2d A 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 A log e A + A - log e A = A
33 3 32 mpancom A log e A + A - log e A = A
34 31 33 eqtr2d A 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 A = log e A + i 2 π A log e A i 2 π n A = log e A + i 2 π n
38 15 34 37 syl2anc A n A = log e A + i 2 π n
39 38 3ad2ant1 A B B 0 n 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 A = log e A + i 2 π n n A = log B + i 2 π n
44 39 43 syl5ibcom A B B 0 e A = B n A = log B + i 2 π n
45 logcl B B 0 log B
46 45 3adant1 A B B 0 log B
47 zcn n n
48 47 adantl A B B 0 n n
49 mulcl i 2 π n i 2 π n
50 20 48 49 sylancr A B B 0 n i 2 π n
51 efadd log B i 2 π n e log B + i 2 π n = e log B e i 2 π n
52 46 50 51 syl2an2r A B B 0 n e log B + i 2 π n = e log B e i 2 π n
53 eflog B B 0 e log B = B
54 53 3adant1 A B B 0 e log B = B
55 ef2kpi n e i 2 π n = 1
56 54 55 oveqan12d A B B 0 n e log B e i 2 π n = B 1
57 simpl2 A B B 0 n B
58 57 mulid1d A B B 0 n B 1 = B
59 52 56 58 3eqtrd A B B 0 n 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 B B 0 n A = log B + i 2 π n e A = B
62 61 rexlimdva A B B 0 n A = log B + i 2 π n e A = B
63 44 62 impbid A B B 0 e A = B n A = log B + i 2 π n