TMUA 2018 Paper 2 Question 19

Divide everything through so each side is a power of the base, and write u = x − y > 0 and v = y − z > 0. The right-hand side of every option is b^u + b^−v, which can be made close to 1 (take u tiny and v huge) or arbitrarily large (take u huge). That kills the three options that compare it with a fixed number 2 or with something bounded above by 2: A has left side 2⋅ 2^z−x < 2, which fails when u is large; B fails when u is large; D fails when u is tiny and v is large. Option C reads 2⋅ 5^u+v > 5^u + 5^−v, and it always holds because 5^u+v > 5^u and 5^u+v > 1 > 5^−v, so adding the two gives the result.