Thank you for your excellent solutions Tan Chor Kiang, age 15, Raffles Institution, Singapore, Robert Goudie, age 16, Madras College, Fife, Scotland and James Selvage, age 17. James' method was very similar to Chor Kiang's, and also involved solving quadratic equations, but James used logarithms.

You have to find all real solutions of the equations
(x2-7x+11)(x2-11x+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 x2-11x+30 = 0.

Conditions: None
x2-11x+30 AMP= 0
(x-6)(x-5) AMP= 0
x - 6 = 0 or x - 5 AMP= 0
x = 6 or x AMP= 5
.

Case 2 x2-7x+11 = 1

Conditions: None
x2-7x+11 AMP= 1
x2-7x+10 AMP= 0
(x-2)(x-5) AMP= 0
x - 2 = 0 or x - 5 AMP= 0
x = 2 or x AMP= 5
.
Case 3 x2-7x+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
x2-7x+11 AMP= -1
x2-7x+12 AMP= 0
(x-3)(x-4) AMP= 0
x - 3 = 0 or x - 4 AMP= 0
x = 3 or x AMP= 4
.
Therefore the possible solutions are x = 2, 3, 4, 5 or 6.