June 2023 Paper 1 Q6
6 Show that the equation
\[2\log_{10} x = \log_{10} 4 + \log_{10}(x + 8)\]has exactly one solution.
Fully justify your answer. [5 marks]
| Scheme | Marks | AO |
|---|---|---|
| Uses power log rule correctly Or Raises 10 to the power of both sides and correctly obtains \(10^{\log_{10} x^2}\) or \(x^2\) PI by correct quadratic | B1 | 1.1b |
| Uses addition or subtraction log rule correctly. Or Correctly combines two indices. PI by correct quadratic. | B1 | 1.1b |
| Solves a three-term quadratic equation obtaining at least one real value for \(x\) | M1 | 1.1a |
| Obtains \(x = 8\) Must have scored B1,B1,M1. | A1 | 1.1b |
| Obtains correct values of \(x\) and explains why \(-4\) is not a solution. Must refer to the log of a negative or state it is only possible to find the log of a positive. Accept correct reference to the domain of a log function. Must have achieved B1,B1,M1,A1 | E1 | 2.4 |
| (5 marks) |
Typical solution
\[2\log_{10} x = \log_{10} 4 + \log_{10}(x + 8)\]\[\log_{10} x^2 = \log_{10} 4(x + 8)\]\[x^2 = 4x + 32\]\[x^2 - 4x - 32 = 0\]\[x = -4 \quad \text{or} \quad 8\]\(-4\) is not a solution as \(\log_{10} -4\) has no real value. Therefore, the equation has exactly one solution.