Use the Index Calculus to solve for $19^x \equiv 205\pmod{337}$, using the factor base $B=\{2,3,5,7\}$ I'm supposed to use the following information to solve the question, but I don't know how.
$$\begin{align}
19^2 &amp;\equiv 2^3 \times 3^1 \times 5^0 \times 7^0&amp;\pmod{337}\\
19^5 &amp;\equiv 2^5 \times 3^0 \times 5^1 \times 7^0&amp;\pmod{337}\\
19^6 &amp;\equiv 2^0 \times 3^0 \times 5^0 \times 7^1&amp;\pmod{337}\\
19^8 &amp;\equiv 2^3 \times 3^1 \times 5^0 \times 7^1&amp;\pmod{337}\\
19^{12} &amp;\equiv 2^0 \times 3^0 \times 5^0 \times 7^2&amp;\pmod{337}\\
19^{18} &amp;\equiv 2^1 \times 3^1 \times 5^0 \times 7^0&amp;\pmod{337}\\
19^{20} &amp;\equiv 2^4 \times 3^2 \times 5^0 \times 7^0&amp;\pmod{337}\\
19^{21} &amp;\equiv 2^3 \times 3^0 \times 5^1 \times 7^0&amp;\pmod{337}\\
19^{24} &amp;\equiv 2^1 \times 3^1 \times 5^0 \times 7^1&amp;\pmod{337}\\
19^{27} &amp;\equiv 2^3 \times 3^0 \times 5^1 \times 7^1&amp;\pmod{337}\\
19^{30} &amp;\equiv 2^1 \times 3^1 \times 5^0 \times 7^2&amp;\pmod{337}\\
19^{36} &amp;\equiv 2^2 \times 3^2 \times 5^0 \times 7^0&amp;\pmod{337}\\
19^{37} &amp;\equiv 2^1 \times 3^0 \times 5^1 \times 7^0&amp;\pmod{337}\\
&amp;\text{and}\\
205 \times 19^{-1} &amp;\equiv 2^6 \times 3^0 \times 5^0 \times 7^0&amp;\pmod{337}\\
205 \times 19^{-5} &amp;\equiv 2^0 \times 3^1 \times 5^2 \times 7^0&amp;\pmod{337}\\
205 \times 19^{-6} &amp;\equiv 2^1 \times 3^3 \times 5^1 \times 7^0&amp;\pmod{337}\\
\end{align}$$
The trick is to take logarithms. The information you have above can be translated as
$$
\begin{align}
2 &amp;= 3\cdot \log_{19} 2 + 1\cdot \log_{19} 3 + 0\cdot \log_{19} 5 + 0\cdot \log_{19} 7 \pmod{336} \\
5 &amp;= 5\cdot \log_{19} 2 + 0\cdot \log_{19} 3 + 1\cdot \log_{19} 5 + 0\cdot \log_{19} 7 \pmod{336} \\
 &amp; \dots
\end{align}
$$
This is basically a linear equation system modulo $336$ (the order of $19$ modulo $337$), at the end of which we obtain the logarithms of $2$, $3$, $5$, and $7$ in base $19$, which are $160$, $194$, $213$, and $6$. (This might require solving the system modulo each factor of $336 = 2^4 \cdot 3 \cdot 7$ individually and recombine with the Chinese remainder theorem).
Having those logarithms, we can use one of the other equations to find the logarithm of $205$. For example,
$$
205\cdot 19^{-1} = 2^6 \cdot 3^0 \cdot 5^0 \cdot 7^0 \pmod{337}
$$
can again be translated as
$$
\log_{19} 205 - 1 = 160\cdot 6 + 194\cdot 0 + 213\cdot 0 + 6\cdot 0 \pmod{336},
$$
from which we easily obtain $\log_{19} 205 = 160\cdot 6 + 1 \bmod 336 = 289$.