Standard +0.3 This is a straightforward logarithm equation requiring systematic application of log laws and exponential manipulation. The steps are routine: divide by 2, exponentiate both sides, rearrange to isolate the exponential term, then take logarithms again. While it requires careful algebraic manipulation across 4 marks, it follows a standard template with no conceptual surprises, making it slightly easier than average.
Remove logarithms and obtain \(5 - e^{-2x} = e^{\frac{1}{2}}\), or equivalent
B1
Obtain a correct value for \(e^{-2x}\), \(e^{2x}\), \(e^{-x}\) or \(e^x\), e.g. \(e^{2x} = 1/(5 - e^{\frac{1}{2}})\)
B1
Use correct method to solve an equation of the form \(e^{2x} = a\), \(e^{-2x} = a\), \(e^x = a\) or \(e^{-x} = a\) where \(a > 0\). [The M1 is dependent on the correct removal of logarithms.]
M1
Obtain answer \(x = -0.605\) only
A1
Total: 4 marks
Remove logarithms and obtain $5 - e^{-2x} = e^{\frac{1}{2}}$, or equivalent | B1 |
Obtain a correct value for $e^{-2x}$, $e^{2x}$, $e^{-x}$ or $e^x$, e.g. $e^{2x} = 1/(5 - e^{\frac{1}{2}})$ | B1 |
Use correct method to solve an equation of the form $e^{2x} = a$, $e^{-2x} = a$, $e^x = a$ or $e^{-x} = a$ where $a > 0$. [The M1 is dependent on the correct removal of logarithms.] | M1 |
Obtain answer $x = -0.605$ only | A1 |
| Total: 4 marks |
---