Moderate -0.3 This is a straightforward numerical methods question requiring rearrangement to f(x)=0 form and application of sign change/interval bisection. The logarithmic manipulation is routine (subtract ln 3, use log laws), and the iterative solution method is standard P3 content with no conceptual challenges beyond careful arithmetic.
Use law of the logarithm of a product, power or quotient or a law of indices (on an expression that is relevant to the question)
M1
e.g. \(\ln(e^{2x}+3)-\ln 3 = \ln\left(\dfrac{e^{2x}+3}{3}\right)\) or \(e^{(2x+\ln 3)}=e^{2x}e^{\ln 3}\)
State a correct equation without logs (in any form)
A1
e.g. \(3+e^{2x}=3e^{2x}\)
Carry out correct method to solve an equation of the form \(e^{2x}=a\), where \(a>0\), or for solving \(e^x=b\) \((b>0)\) if they have already taken the square root
M1
Allow for \(x=\frac{1}{2}\ln\frac{3}{2}\). M1 can be implied by correct answer.
Obtain answer \(x=0.203\)
A1
CAO. The question requires 3 d.p. Answer only with no working shown is 0/4.
Total
4
**Question 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use law of the logarithm of a product, power or quotient or a law of indices (on an expression that is relevant to the question) | M1 | e.g. $\ln(e^{2x}+3)-\ln 3 = \ln\left(\dfrac{e^{2x}+3}{3}\right)$ or $e^{(2x+\ln 3)}=e^{2x}e^{\ln 3}$ |
| State a correct equation without logs (in any form) | A1 | e.g. $3+e^{2x}=3e^{2x}$ |
| Carry out correct method to solve an equation of the form $e^{2x}=a$, where $a>0$, or for solving $e^x=b$ $(b>0)$ if they have already taken the square root | M1 | Allow for $x=\frac{1}{2}\ln\frac{3}{2}$. M1 can be implied by correct answer. |
| Obtain answer $x=0.203$ | A1 | CAO. The question requires 3 d.p. Answer only with no working shown is 0/4. |
| **Total** | **4** | |