Moderate -0.5 This is a straightforward application of the logarithm subtraction law requiring rearrangement to ln((x+5)/x) = 5, then exponentiating and solving a linear equation. It's slightly easier than average as it involves only one standard technique with no conceptual difficulty, though the algebraic manipulation after exponentiating requires care.
Use exponentials or law for the logarithm of a product, quotient or power
\(M1^*\)
\(e^{\ln(5+x)} = e^{5+\ln x}\) insufficient. Need e.g. \(\ln\left(\frac{x+5}{x}\right) = 5\) or \(\ln(x+5) = \ln(e^5) + \ln x\) or \(\ln(x+5) = \ln(e^5 x)\) or \(x + 5 = e^{5+\ln x}\) or \(x + 5 = e^5 e^{\ln x}\) and others.
Correctly remove logarithms
\(DM1\)
Obtain a correct equation in \(x\)
\(A1\)
e.g. \(\frac{x+5}{x} = e^5\) (or \(148.4...\)) or \(x + 5 = xe^5\)
Obtain \(0.034\)
\(A1\)
CAO Final answer must be 3d.p.
4
**Question 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use exponentials or law for the logarithm of a product, quotient or power | $M1^*$ | $e^{\ln(5+x)} = e^{5+\ln x}$ insufficient. Need e.g. $\ln\left(\frac{x+5}{x}\right) = 5$ or $\ln(x+5) = \ln(e^5) + \ln x$ or $\ln(x+5) = \ln(e^5 x)$ or $x + 5 = e^{5+\ln x}$ or $x + 5 = e^5 e^{\ln x}$ and others. |
| Correctly remove logarithms | $DM1$ | |
| Obtain a correct equation in $x$ | $A1$ | e.g. $\frac{x+5}{x} = e^5$ (or $148.4...$) or $x + 5 = xe^5$ |
| Obtain $0.034$ | $A1$ | CAO Final answer must be 3d.p. |
| | **4** | |