Dgdrxrt

I'm not sure how to start this, I know I'm supposed to use log to help me out..But my textbook isn't very clear

You need a primitive root of $37$ to start. Fortunately, the first thing to try is $2$ and it works. Second, work out a table of indices to the base $2$ modulo $37$.

You'll discover that $mboxind_2 3 = 26$ and $mboxind_2 11 = 30.$ So your congruence becomes

$$left(2^26right)^xequiv 2^30 pmod37.$$

Which means

$$26x equiv 30 pmodphi(37).$$

You can divide through by $2$ to get

$$13x equiv 15 pmod18.$$

The solutions of this congruence are your answers.

Computing the first few values of $3^nbmod 37$, you'll find the order of $3mod 37$, which must be a divisor of $36$ anyway, by Fermat's and Lagrange's theorems. It is not hard to find that
$$3^3equiv -10,quad3^6equiv 100equiv-11,quad 3^9=110equiv -1,quad 3^18equiv 1mod 37.$$
So the order of $3$ modulo $37$ is equal to $18$.

Furthermore, these computations yield a solution: $x=15$, since
$$11=(-1)(-11)equiv3^6cdot 3^9=3^15.$$
Now, if $x$ is another solution, we have $3^xequiv 3^15$, whence $3^x-15equiv 1mod 37$, so
$$x-15equiv 0mod 18,quadtextwhence ;xequiv 15mod 18.$$

I can turn this into a slightly weird statement via the following: $$begineqnarray3^xequiv 11 pmod 37text (explored mod 2, reveals even multiplier)\3^xequiv11pmod 74text (explored mod 3, reveals 2 mod 3 multiplier )\3^xequiv 159pmod222text (implies a relationship among primes when reduced)\3^x-1equiv 53 pmod 222endeqnarray$$

for this last one we can test x-1 for being even (53 being a quadratic residue mod 222 testing via factors 37,2,3) https://en.m.wikipedia.org/wiki/Quadratic_reciprocity tells us that 53 is a quadratic residue mod 37 but a quadratic non-residue mod 3, and it's a quadratic residue mod 2, So sadly this doesn't quite help us. Since 53 isn't a direct power of 3, x is at least 5 because that's the first time it exceeds our modulus and create a non-power remainder. 1 mod 5 multipliers are out, because they create a multiple of 5 being a power of 3, erroneously (275 mod 1110 in fact). I've removed a third of your work without logs being used much. Literally now apply log rules to figure it out after that. EDIT okay $$3^x-1equiv 53bmod 74$$ actually doh so it is a quadratic residue) meaning x is odd, we still have x>3 though so at least 5 is still justified. we just know it's an odd exponent now. disallowed multipliers go to 3 mod 5 instead when brought properly back to mod 222.

deeper explanation:
$$begineqnarraycequiv bbmod aimplies c=ay+b\37equiv 1bmod 2\74equiv 0bmod 2\74equiv2bmod 3\11equiv1bmod 2\11equiv 2bmod 3\3^xequiv 1bmod 2\3^xequiv 0 bmod 3endeqnarray$$

With the above the first turns into:

$$1equiv y+1 bmod 2$$ forcing y to be 0 mod 2. This means it's $$74z +11$$ Modding both sides of the second above gives:

$$0equiv 2z+2bmod 3$$ which solves for z being -1 ( aka 2) mod 3 $$begineqnarray222=3cdot74\159=2cdot74+11endeqnarray$$

dividing 222, 159, and $3^x$ by 3 lowers it to: $$3^x-1equiv 53bmod 74$$ which you can show is a quadratic residue as $$begineqnarray53equiv 1^2bmod 2\53equiv (pm4)^2 bmod 37endeqnarray$$ and apply chinese remainder theorem to get to it's mod 74 equivalents of $$(pm33)^2 bmod 74$$ now to find an order mod 74 ( or the equivalents mod 222) to find out more. I simply converted back to mod 222 before, and because $$222>81=3^4$$ and x-1 producing a quadratic residue, we know x is odd and at least 5. Oh, and since all x values that work need be odd you get for free the order of 1 needs be even( fitting their difference). Then you can apply other things to narrow the repeat length down to 6,12,18, or 36.