Higher June 2023 Paper 1R Q24

24 Given that

\(2^n = 2^{x^2} \times 16^x \times 8\)

and

\(x \gt 0\)

find an expression for \(x\) in terms of \(n\)
State any restrictions on \(n\)

(5)

Mark scheme

Scheme	Marks
\(2^3\) and \(2^{4x}\) or \((2^4)^x\)	M1
\(n = x^2 + 4x + 3\) oe or \(x^2 + 4x + 3 - n = 0\)	A1
\((n =)\;(x + 2)^2 - 2^2\)....... oe or \((x =)\;-2 \pm \sqrt{n + 1}\) \((x =)\;\dfrac{-4 \pm \sqrt{4^2 - 4 \times 1 \times (3 - n)}}{2}\) oe	M1
\((x =)\;-2 + \sqrt{n + 1}\) oe or \((x =)\;\dfrac{-4 + \sqrt{4^2 - 4 \times 1 \times (3 - n)}}{2}\) oe	A1
Correct answer scores full marks (unless from obvious incorrect working) Answer: \((x =)\;-2 + \sqrt{n + 1}\) and \(n \gt 3\)	A1
	(5)
	(5 marks)

M1: for writing \(16^x\) and 8 as a power of 2 (or all as powers of 4,8 or 16)

A1: for writing \(n\) in terms of \(x\)
correct expression implies first M1

M1: for a correct first step in completing the square or using the quadratic formula correctly ft their 3 term quadratic

A1: for correctly rearranging to make \(x\) the subject (must be positive square root)

A1: must be positive square root

Accept \((x =)\;\sqrt{n + 1} - 2\) oe and \(3 \lt n\)

\((x =)\;\dfrac{-4 + \sqrt{4^2 - 4 \times 1 \times (3 - n)}}{2}\) oe

and \(n \gt 3\) or \(3 \lt n\)

Mark scheme (ALT)

Scheme	Marks
\(4^{\frac{1}{2}n}\), \(4^{\frac{1}{2}x^2}\), \(4^{2x}\) and \(4^{\frac{3}{2}}\) or \(8^{\frac{1}{3}n}\), \(8^{\frac{1}{3}x^2}\) and \(8^{\frac{4}{3}x}\) or \(16^{\frac{1}{4}n}\), \(16^{\frac{1}{4}x^2}\) and \(16^{\frac{3}{4}}\)	M1
\(n = x^2 + 4x + 3\) oe or \(x^2 + 4x + 3 - n = 0\)	A1
\((n =)\;(x + 2)^2 - 2^2\)....... oe or \((x =)\;-2 \pm \sqrt{n + 1}\) \((x =)\;\dfrac{-4 \pm \sqrt{4^2 - 4 \times 1 \times (3 - n)}}{2}\) oe	M1
\((x =)\;-2 + \sqrt{n + 1}\) oe or \((x =)\;\dfrac{-4 + \sqrt{4^2 - 4 \times 1 \times (3 - n)}}{2}\) oe	A1
Correct answer scores full marks (unless from obvious incorrect working) Answer: \((x =)\;-2 + \sqrt{n + 1}\) and \(n \gt 3\)	A1