Solving $7^{k}\; \equiv\; 11\; \left( \mbox{mod}\; 26 \right)$

Question

How do you solve modular equations where the variable is in the exponent like:

$7^{k}\; \equiv\; 11\; \left( \mbox{mod}\; 26 \right)$

Do you have to use fermats little theorem or euler totient function or something of that sort?

Answer 1

By inspection, $\displaystyle7^2\equiv-3,7^3\equiv5\implies7^5\equiv-15\equiv11\pmod{26}\implies 7^k\equiv 7^5\iff 7^{k-5}\equiv1$

Now, $\displaystyle7^6\equiv25\equiv-1\implies7^{12}\equiv1\implies$ord$_{26}7=12$

$\displaystyle\implies 12|(k-5)\iff k\equiv5\pmod{12}$

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Alternatively, observe that ord$\displaystyle_{26}7=12=\phi(26)\implies 7$ is a primitive root $\pmod{26}$

So applying Discrete Logarithm on $$7^k\equiv11\equiv7^5\pmod{26}$$

we get, $$k\equiv 5\pmod{\phi(26)}$$

Answer 2

How do you solve modular equations where the variable is in the exponent?

All you need to know is that there are only n possible remainders modulo n, so $a^x\text{mod }n$ can only have n possible outcomes, thus inevitably generating a cycle of length $\le n$. So all you need is just a bit of patience. And eventually a computer. :-)