Hint
The substitution \(\displaystyle u=a^{x}\) requires u>0. Factor the quadratic in the positive variable \(\displaystyle a^{x}\).
Worked solution
(a)
Put U=2ˣ>0. Then \(\left(U - 32\right) \left(U - 2\right)\)=0; the two positive powers give the stated exponents.
Answer: \(\left\{1, 5\right\}\)
(b)
Recover x from the second positive power.
\(5\)
Answer: \(5\)
(c)
Multiply the recovered exponents, not the substituted roots.
\(5\)
Answer: \(5\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternatives and consequential reasoning; an unexplained final answer does not earn a method mark.
(a) · 2 marks
- 1 mark: Substitute a positive exponential variable, solve its quadratic and reverse the substitution.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: A root in u is not yet an answer for x.
(b) · 2 marks
- 1 mark: Recover x from the second positive power.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Do not accept a negative value for \(\displaystyle a^{x}\).
(c) · 2 marks
- 1 mark: Multiply the recovered exponents, not the substituted roots.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: A root in u is not yet an answer for x.
Total: 6 marks.