Moderate -0.8 This is a straightforward single-step natural logarithm equation requiring only routine manipulation: exponentiate both sides to get 1 + e^(-3x) = e^2, rearrange to isolate the exponential term, take ln again, and solve for x. It's a standard textbook exercise with no conceptual difficulty or problem-solving required, making it easier than average but not trivial since it involves exponential/logarithm manipulation.
Use correct method to solve an equation of the form \(e^{-3x} = a\), where \(a > 0\), for \(x\) or equivalent
M1
(\(e^{-3x} = 6.389...\)) Evidence of method must be seen
Obtain answer \(x = -0.618\) only
A1
Must be 3 decimal places
Alternative method for question 1:
Answer
Marks
Guidance
Answer
Marks
Guidance
State that \(1 + e^{-3x} = e^2\)
B1
Rearrange to obtain an expression for \(e^x\) and solve an equation of the form \(e^x = a\), where \(a > 0\), or equivalent
M1
\(e^x = \sqrt[3]{\dfrac{1}{e^2 - 1}}\)
Obtain answer \(x = -0.618\) only
A1
Must be 3 decimal places
Total
3
## Question 1:
| Answer | Marks | Guidance |
|--------|-------|----------|
| State that $1 + e^{-3x} = e^2$ | **B1** | With no errors seen to that point |
| Use correct method to solve an equation of the form $e^{-3x} = a$, where $a > 0$, for $x$ or equivalent | **M1** | ($e^{-3x} = 6.389...$) Evidence of method must be seen |
| Obtain answer $x = -0.618$ only | **A1** | Must be 3 decimal places |
**Alternative method for question 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| State that $1 + e^{-3x} = e^2$ | **B1** | |
| Rearrange to obtain an expression for $e^x$ and solve an equation of the form $e^x = a$, where $a > 0$, or equivalent | **M1** | $e^x = \sqrt[3]{\dfrac{1}{e^2 - 1}}$ |
| Obtain answer $x = -0.618$ only | **A1** | Must be 3 decimal places |
| **Total** | **3** | |
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