Standard +0.3 This is a straightforward exponential equation that becomes a simple quadratic after substitution u = 3^x, leading to u² - 2u - 1 = 0. The solution requires only standard A-level techniques (substitution, quadratic formula, logarithms) with no novel insight needed. It's slightly easier than average as the substitution is explicitly given and the algebraic manipulation is routine.
State or imply at any stage that \(3^{-x} = \frac{1}{3^x}\), or that \(3^{-x} = \frac{1}{u}\) where \(u = 3^x\)
B1
Convert given equation into the 3-term quadratic in \(u\) (or \(3^x\)): \(u^2 - 2u - 1 = 0\)
B1
Solve a 3-term quadratic, obtaining one or two roots
M1
Obtain root \(\frac{2 + \sqrt{8}}{2}\), or a simpler equivalent, or decimal value in [2.40, 2.42]
A1
Use a correct method for finding the value of \(x\) from a positive root
M1
Obtain \(x = 0.802\) only
A1
Total: 6 marks
State or imply at any stage that $3^{-x} = \frac{1}{3^x}$, or that $3^{-x} = \frac{1}{u}$ where $u = 3^x$ | B1 |
Convert given equation into the 3-term quadratic in $u$ (or $3^x$): $u^2 - 2u - 1 = 0$ | B1 |
Solve a 3-term quadratic, obtaining one or two roots | M1 |
Obtain root $\frac{2 + \sqrt{8}}{2}$, or a simpler equivalent, or decimal value in [2.40, 2.42] | A1 |
Use a correct method for finding the value of $x$ from a positive root | M1 |
Obtain $x = 0.802$ only | A1 | **Total: 6 marks**
---