June 2024 Paper 1 Q21
21
\[\left(2^{x + 3}\right)^{4x + 1} = 8^{x - 1}\]Work out the possible values of \(x\). [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(2^{(4x + 1)(x + 3)}\) or \(2^{4x^2 + 13x + 3}\) | M1 | lhs |
| \(\left(2^3\right)^{x - 1}\) or \(2^{3(x - 1)}\) or \(2^{3x - 3}\) | M1 | rhs |
| \(4x^2 + 12x + x + 3 = 3x - 3\) or \(4x^2 + 10x + 6\ (= 0)\) or \(2x^2 + 5x + 3\ (= 0)\) | M1dep | oe with expanded brackets dependent on both M marks \(4x^2 + 12x + x + 3\) implies first M mark \(3x - 3\) implies second M mark |
| \((2x + 3)(x + 1)\ (= 0)\) or if quadratic formula used \(\dfrac{-5 \pm \sqrt{5^2 - 24}}{4}\) or better | M1dep | oe factorised |
| \(-1\) and \(-1\dfrac{1}{2}\) or \(\dfrac{-3}{2}\) or \(-1.5\) | A1 |
Additional guidance
| Both answers correct without working will get all marks | M4A1 |
| Only one solution correct without working scores nothing | M0A0 |
| No oe for A mark | |
| Possible to make both sides powers of 8. Mark to the equivalent of this MS |