| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Simple exponential equation solving |
| Difficulty | Standard +0.3 This is a standard C2 exponential question with routine techniques: sketching an exponential curve (trivial), solving 9^x = 150 using logarithms (straightforward), and algebraic manipulation to reach a given form. The final part requires more algebraic steps but follows a predictable pattern for this topic, making it slightly above average difficulty for C2 but still routine. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | [Graph shown] | M1, A1 |
| B1 | State or imply (0, 6) | |
| (ii) | \(9^x = 150\) | M1 |
| \(x\log 9 = \log 150\) | M1 | Use \(\log a^b = b\log a\) and attempt correct method to find x |
| \(x = 2.28\) | A1 | Obtain \(x = 2.28\) |
| (iii) | \(6 \times 5^x = 9^x\) | M1 |
| \(\log_3(6 \times 5^x) = \log_3 9^x\) | M1 | Use \(\log a^b = b\log a\) correctly on \(\log 5^x\) or \(\log 9^x\) or legitimate combination of these two |
| \(\log_3 6 + x\log_3 5 = x\log_3 9\) | M1 | Use \(\log ab = \log a + \log b\) correctly on \(\log(6 \times 5^x)\) or log 6 |
| \(\log_3 3 + \log_3 2 + x\log_3 5 = 2x\) | M1 | Use \(\log_3 9 = 2\) or equiv (need base 3 throughout that line) |
| \(x(2 - \log_3 5) = 1 + \log_3 2\) | ||
| \(x = \frac{1 + \log_3 2}{2 - \log_3 5}\) A.G. | A1 | Obtain \(x = \frac{1+\log_3 2}{2-\log_3 5}\) convincingly (inc base 3 throughout) |
(i) | [Graph shown] | M1, A1 | Reasonable graph in both quadrants; Correct graph in both quadrants |
| | B1 | State or imply (0, 6) |
(ii) | $9^x = 150$ | M1 | Introduce logarithms throughout, or equiv with $\log_9$ |
| $x\log 9 = \log 150$ | M1 | Use $\log a^b = b\log a$ and attempt correct method to find x |
| $x = 2.28$ | A1 | Obtain $x = 2.28$ |
(iii) | $6 \times 5^x = 9^x$ | M1 | Form eqn in x and take logs throughout (any base) |
| $\log_3(6 \times 5^x) = \log_3 9^x$ | M1 | Use $\log a^b = b\log a$ correctly on $\log 5^x$ or $\log 9^x$ or legitimate combination of these two |
| $\log_3 6 + x\log_3 5 = x\log_3 9$ | M1 | Use $\log ab = \log a + \log b$ correctly on $\log(6 \times 5^x)$ or log 6 |
| $\log_3 3 + \log_3 2 + x\log_3 5 = 2x$ | M1 | Use $\log_3 9 = 2$ or equiv (need base 3 throughout that line) |
| $x(2 - \log_3 5) = 1 + \log_3 2$ | | |
| $x = \frac{1 + \log_3 2}{2 - \log_3 5}$ A.G. | A1 | Obtain $x = \frac{1+\log_3 2}{2-\log_3 5}$ convincingly (inc base 3 throughout) |9 (i) Sketch the curve $y = 6 \times 5 ^ { x }$, stating the coordinates of any points of intersection with the axes.\\
(ii) The point $P$ on the curve $y = 9 ^ { x }$ has $y$-coordinate equal to 150 . Use logarithms to find the $x$-coordinate of $P$, correct to 3 significant figures.\\
(iii) The curves $y = 6 \times 5 ^ { x }$ and $y = 9 ^ { x }$ intersect at the point $Q$. Show that the $x$-coordinate of $Q$ can be written as $x = \frac { 1 + \log _ { 3 } 2 } { 2 - \log _ { 3 } 5 }$.
\hfill \mbox{\textit{OCR C2 2010 Q9 [11]}}