June 2022 Paper 2 Q22
22 Using powers of 2 or otherwise, work out the non-zero value of \(x\) for which
\[\left(16^x\right)^x = \dfrac{1}{2^{3x}}\]You must show your working. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Uses powers of 2 | ||
| \((16^x =)\; 2^{4x}\) or \(((16^x)^x =)\; (2^4)^{x^2}\) | M1 | implied by \(((16^x)^x =)\; 2^{4x^2}\) may be implied by 3rd M1 |
| \(((16^x)^x =)\; 2^{4x^2}\) | M1dep | implied by \(2^{4x^2 + 3x}\) may be implied by 3rd M1 |
| Correct quadratic equation or correct linear equation or correct equation involving indices with the same base | M1dep | eg \(4x^2 = -3x\) or \(4x^2 + 3x = 0\) or \(4x = -3\) or \(2^{4x^2} = 2^{-3x}\) or \(2^{4x^2 + 3x} = 2^0\) do not allow if the equation is from incorrect working do not allow if the only equation is \(x = -\dfrac{3}{4}\) |
| M3 and \(-\dfrac{3}{4}\) | A1 | oe ignore inclusion of answer 0 |
| Alternative method 2 Uses powers of 16 | ||
| \(((16^x)^x =)\; 16^{x^2}\) or \(\left(\dfrac{1}{2^{3x}} =\right) \dfrac{1}{\left(16^{\frac{1}{4}}\right)^{3x}}\) | M1 | implied by \(\left(\dfrac{1}{2^{3x}} =\right) \dfrac{1}{16^{\frac{3x}{4}}}\) or \(\left(\dfrac{1}{2^{3x}} =\right) 16^{-\frac{3x}{4}}\) may be implied by 3rd M1 |
| \(((16^x)^x =)\; 16^{x^2}\) and \(\left(\dfrac{1}{2^{3x}} =\right) \dfrac{1}{16^{\frac{3x}{4}}}\) | M1dep | oe eg \(((16^x)^x =)\; 16^{x^2}\) and \(\left(\dfrac{1}{2^{3x}} =\right) 16^{-\frac{3x}{4}}\) may be implied by 3rd M1 |
| Correct quadratic equation or correct linear equation or correct equation involving indices with the same base | M1dep | eg \(x^2 = -\dfrac{3}{4}x\) or \(4x^2 + 3x = 0\) or \(16^{x^2} = 16^{-\frac{3x}{4}}\) do not allow if the equation is from incorrect working do not allow if the only equation is \(x = -\dfrac{3}{4}\) |
| M3 and \(-\dfrac{3}{4}\) | A1 | oe ignore inclusion of answer 0 |
| Alternative method 3 Uses powers of 4 | ||
| \((16^x =)\; 4^{2x}\) or \(((16^x)^x =)\; (4^2)^{x^2}\) or \(\left(\dfrac{1}{2^{3x}} =\right) \dfrac{1}{\left(4^{\frac{1}{2}}\right)^{3x}}\) | M1 | implied by \(((16^x)^x =)\; 4^{2x^2}\) or \(\left(\dfrac{1}{2^{3x}} =\right) \dfrac{1}{4^{\frac{3x}{2}}}\) or \(\left(\dfrac{1}{2^{3x}} =\right) 4^{-\frac{3}{2}x}\) may be implied by 3rd M1 |
| \(((16^x)^x =)\; 4^{2x^2}\) and \(\left(\dfrac{1}{2^{3x}} =\right) \dfrac{1}{4^{\frac{3x}{2}}}\) | M1dep | oe \(((16^x)^x =)\; 4^{2x^2}\) and \(\left(\dfrac{1}{2^{3x}} =\right) 4^{-\frac{3}{2}x}\) may be implied by 3rd M1 |
| Correct quadratic equation or correct linear equation or correct equation involving indices with the same base | M1dep | eg \(2x^2 = -\dfrac{3}{2}x\) or \(4x^2 + 3x = 0\) or \(4^{2x^2} = 4^{-\frac{3}{2}x}\) do not allow if the equation is from incorrect working do not allow if the only equation is \(x = -\dfrac{3}{4}\) |
| M3 and \(-\dfrac{3}{4}\) | A1 | oe ignore inclusion of answer 0 |
| Alternative method 4 Takes the \(x\)th root of each side and uses powers of 2 | ||
| \((16^x =)\; 2^{4x}\) or \(16^x = \left(\dfrac{1}{2^{3x}}\right)^{\frac{1}{x}}\) | M1 | oe eg \(16^x = \sqrt[x]{\dfrac{1}{2^{3x}}}\) or \(16^x = \dfrac{1}{2^3}\) or \(16^x = 2^{-3}\) may be implied by 3rd M1 |
| \(2^{4x} = \left(\dfrac{1}{2^{3x}}\right)^{\frac{1}{x}}\) | M1dep | oe eg \(2^{4x} = \dfrac{1}{2^3}\) may be implied by 3rd M1 |
| Correct quadratic equation or correct linear equation or correct equation involving indices with the same base | M1dep | eg \(4x = -3\) or \(2^{4x} = 2^{-3}\) do not allow if the equation is from incorrect working do not allow if the only equation is \(x = -\dfrac{3}{4}\) |
| M3 and \(-\dfrac{3}{4}\) | A1 | oe ignore inclusion of answer 0 |
Additional guidance
| Up to M2 may be awarded for correct work with no, or incorrect answer, even if this is seen amongst multiple attempts | |
| Allow \(2^{4 \times x \times x}\) for \(2^{4x^2}\) etc | |
| Responses using other powers eg powers of 8 can be escalated | Escalate |
| Ignore simplification or conversion if correct answer seen |