Solution

Use the table of powers and indexes provided to solve the following congruence: $13^x \equiv 6 \pmod{23}$

Some powers and related indexes $\pmod{23}$
$$\begin{array}{rclcrcl|rclrcl} 5^1 &\equiv& 5 && 5^{12} &\equiv& 18 & \text{ind}_5 1 &=& 22 & \text{ind}_5 12 &=&20\\ 5^2 &\equiv& 2 && 5^{13} &\equiv& 21 & \text{ind}_5 2 &=& 2 & \text{ind}_5 13 &=&14\\ 5^3 &\equiv& 10 && 5^{14} &\equiv& 13 & \text{ind}_5 3 &=& 16 & \text{ind}_5 14 &=&21\\ 5^4 &\equiv& 4 && 5^{15} &\equiv& 19 & \text{ind}_5 4 &=& 4 & \text{ind}_5 15 &=&17\\ 5^5 &\equiv& 20 && 5^{16} &\equiv& 3 & \text{ind}_5 5 &=& 1 & \text{ind}_5 16 &=&8\\ 5^6 &\equiv& 8 && 5^{17} &\equiv& 15 & \text{ind}_5 6 &=& 18 & \text{ind}_5 17 &=&7\\ 5^7 &\equiv& 17 && 5^{18} &\equiv& 6 & \text{ind}_5 7 &=& 19 & \text{ind}_5 18 &=&12\\ 5^8 &\equiv& 16 && 5^{19} &\equiv& 7 & \text{ind}_5 8 &=& 6 & \text{ind}_5 19 &=&15\\ 5^9 &\equiv& 11 && 5^{20} &\equiv& 12 & \text{ind}_5 9 &=& 10 & \text{ind}_5 20 &=&5\\ 5^{10} &\equiv& 9 && 5^{21} &\equiv& 14 & \text{ind}_5 10 &=& 3 & \text{ind}_5 21 &=&13\\ 5^{11} &\equiv& 22 && 5^{22} &\equiv& 1 & \text{ind}_5 11 &=& 9 & \text{ind}_5 22 &=&11\\ \end{array}$$

Starting with

$$13^x \equiv 6 \pmod{23}$$

we take index base $5$ of both sides

$$\text{ind}_5 13^x \equiv \text{ind}_5 6 \pmod{22}$$

so that we can use the properties of index arithmetic to pull the exponent down as a multiplier

$$x \cdot \text{ind}_5 13 \equiv \text{ind}_5 6 \pmod{22}$$

This produces a linear congruence, which should be more familiar...

$$14x \equiv 18 \pmod{22}$$

Since $\gcd(14,22) = 2$, we recall that division of both sides by $2$ will require dividing the modulus by $2$

$$7x \equiv 9 \pmod{11}$$

Finding $7^{-1} \equiv 8 \pmod{11}$ in the usual way (i.e., via the Euclidean Algorithm), we multiply both sides by this value to obtain

$$x \equiv 6 \pmod{11}$$

Finally, by considering what remainders $\pmod{22}$ are congruent to $6 \pmod{11}$, we arrive at the solution:

$$x \equiv 6 \text{ or } 17 \pmod{22}$$