What's a half on the clock? In my room, I have this geeky clock (click for full size):

Most of these are not difficult to figure out, but the one for 4-o-clock is particularly tricky:

Normally, a fraction like 1/2 doesn't make sense in modular arithmetic since only integers are involved. The correct way, then, is to see this as the inverse of 2, or to put it another way,  is that number  where . Put this way, a moment's thought will reveal that  because .
However, simply finding the multiplicative inverse would be far too easy as a challenge. So let's bump up the difficulty to exponentiation, or in other words, finding the modular logarithm or discrete logarithm of 2. In this case, 3 is the modular logarithm of 2 with respect to 7. For those of you with number theory/abstract algebra background, this means calculating the multiplicative order of 2 modulo n.
The Challenge
Given a positive odd integer n greater than 1, output the smallest positive integer x where .
Examples
n x
3 2 
5 4 
7 3 
9 6 
11 10 
13 12 
15 4 
17 8 
19 18 
21 6 
23 11 
25 20 
27 18 
29 28 
31 5 
33 10 
35 12 
37 36 
39 12 
41 20 
43 14 
45 12 
47 23 
49 21 
51 8 
53 52 
55 20 
57 18 
59 58 
61 60 
63 6 
65 12 
67 66 
69 22 
71 35 
73 9 
75 20 
77 30 
79 39 
81 54 
83 82 
85 8 
87 28 
89 11 
91 12 
93 10 
95 36 
97 48 
99 30 
101 100 
103 51 
105 12 
107 106 
109 36 
111 36 
113 28 
115 44 
117 12 
119 24 
121 110 
123 20 
125 100 
127 7 
129 14 
131 130 
133 18 
135 36 
137 68 
139 138 
141 46 
143 60 
145 28 
147 42 
149 148 
151 15 
153 24 
155 20 
157 52 
159 52 
161 33 
163 162 
165 20 
167 83 
169 156 
171 18 
173 172 
175 60 
177 58 
179 178 
181 180 
183 60 
185 36 
187 40 
189 18 
191 95 
193 96 
195 12 
197 196 
199 99 
201 66
Jelly, 6 bytes

R2*%i1


Try it online!

How it works

R2*%i1  Input: n

R       Range; yield [1, ..., n].
 2*     Compute [2**1, ..., 2**n].
   %    Hook; take all powers modulo n.
    i1  Get the index of the first 1.
        Indices are 1-based, so index k corresponds to the natural number k.