Thank you for your excellent solutions Tan Chor Kiang, Raffles Institution, Singapore, Robert Goudie, Madras College, Fife, Scotland and James. Chor Kiang's and James' methods were very similar, and involved solving quadratic equations, but James used logarithms.

You have to find all real solutions of the equations $$(x^2-7x+11)^{(x^2-11x+30)} = 1.$$ This is Chor Kiang's solution: The left hand side can only be 1 if the base is 1, or -1 or the index is 0.

Hence, you have 3 cases.

Case 1 $x^2-11x+30 = 0$.

Conditions: None

\begin{eqnarray}x^2-11x+30 &=& 0 \\ (x-6)(x-5) &=& 0 \\ x - 6 &=& 0\; \mathrm{or} \; x-5&=&0\\ x &=& 6\; \mathrm{or} \; x &=& 5\end{eqnarray}

Case 2 $x^2-7x+11 = 1$

Conditions: None

\begin{eqnarray} x^2-7x+11 &=& 1 \\ x^2-7x+10 &=& 0 \\ (x-2)(x-5) &=& 0 \\ x - 2 &=& 0 \; \mathrm{or} \; x - 5 &=& 0 \\ x &=& 2\; \mathrm{or} \; x&=&5. \end{eqnarray}

Case 3 $x^2-7x+11 = -1$

Conditions: Index must be an even number. Proof: This is already true, since the index is $(x-6)(x-5)$. Hence, one of the numbers must be an even number. When and even number multiplies an odd number, an even number results

\begin{eqnarray} x^2-7x+11 &=& -1 \\ x^2-7x+12 &=& 0 \\ (x-3)(x-4) &=& 0 \\ x - 3 &=& 0\; \mathrm{or} \; x - 4 &=& 0 \\ x &=& 3\; \mathrm{or} \; x &=& 4. \end{eqnarray}
Therefore the possible solutions are $x = 2, \; 3, \; 4, \; 5$ or $6$.