| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2002 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Quadratic in exponential form |
| Difficulty | Moderate -0.8 This is a standard substitution problem requiring recognition that 9^x = (3^2)^x = (3^x)^2 = y^2, leading to a simple quadratic equation 2y^2 - 7y + 3 = 0. The mechanics are routine: factorize or use quadratic formula, then take logarithms. This is easier than average as it's a textbook exercise with a clear method and no conceptual obstacles. |
| Spec | 1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State or imply that \(9^x = y^3\) | B1 | 1 |
| (ii) Carry out recognisable solution method for quadratic in \(y\) | M1 | |
| Obtain \(y = \frac{1}{3}\) and \(y = 3\) from \(2y^2 - 7y + 3 = 0\) | A1 | |
| Use log method to solve an equation of the form \(3^x = k\) | M1 | |
| Obtain answer \(x = \frac{\ln 2}{\ln 3}\), or exact equivalent | A1 | ⊕ |
| State exact answer \(x = 1\) (no penalty if logs used) | B1 | 5 |
**(i)** State or imply that $9^x = y^3$ | B1 | 1 |
**(ii)** Carry out recognisable solution method for quadratic in $y$ | M1 |
Obtain $y = \frac{1}{3}$ and $y = 3$ from $2y^2 - 7y + 3 = 0$ | A1 |
Use log method to solve an equation of the form $3^x = k$ | M1 |
Obtain answer $x = \frac{\ln 2}{\ln 3}$, or exact equivalent | A1 | ⊕ |
State exact answer $x = 1$ (no penalty if logs used) | B1 | 5 |
---3 (i) Express $9 ^ { x }$ in terms of $y$, where $y = 3 ^ { x }$.\\
(ii) Hence solve the equation
$$2 \left( 9 ^ { x } \right) - 7 \left( 3 ^ { x } \right) + 3 = 0 ,$$
expressing your answers for $x$ in terms of logarithms where appropriate.
\hfill \mbox{\textit{CAIE P2 2002 Q3 [6]}}