Standard +0.8 This question requires applying logarithm laws to form a quadratic, solving it, then critically checking domain restrictions (x > 0) and rejecting an extraneous solution. The 'show exactly one solution' with 'fully justify' demands rigorous reasoning beyond routine manipulation, making it moderately challenging but still within standard A-level scope.
\(2\log_{10} x = \log_{10} 4 + \log_{10}(x+8)\) → uses power log rule; or raises 10 to the power of both sides to get \(10^{\log_{10} x^2}\) or \(x^2\)
B1
PI by correct quadratic
\(\log_{10} x^2 = \log_{10} 4(x+8)\) → uses addition/subtraction log rule correctly, or correctly combines two indices
B1
PI by correct quadratic
\(x^2 = 4x + 32\), so \(x^2 - 4x - 32 = 0\) → solves three-term quadratic obtaining at least one real value for \(x\)
M1
\(x = 8\)
A1
Must have scored B1, B1, M1
\(x = -4\) is not a solution as \(\log_{10}(-4)\) has no real value; therefore the equation has exactly one solution
E1
Must refer to log of a negative or state it is only possible to find the log of a positive; accept correct reference to domain of a log function; must have achieved B1,B1,M1,A1
## Question 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2\log_{10} x = \log_{10} 4 + \log_{10}(x+8)$ → uses power log rule; or raises 10 to the power of both sides to get $10^{\log_{10} x^2}$ or $x^2$ | B1 | PI by correct quadratic |
| $\log_{10} x^2 = \log_{10} 4(x+8)$ → uses addition/subtraction log rule correctly, or correctly combines two indices | B1 | PI by correct quadratic |
| $x^2 = 4x + 32$, so $x^2 - 4x - 32 = 0$ → solves three-term quadratic obtaining at least one real value for $x$ | M1 | |
| $x = 8$ | A1 | Must have scored B1, B1, M1 |
| $x = -4$ is not a solution as $\log_{10}(-4)$ has no real value; therefore the equation has exactly one solution | E1 | Must refer to log of a negative or state it is only possible to find the log of a positive; accept correct reference to domain of a log function; must have achieved B1,B1,M1,A1 |
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