| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indices and Surds |
| Type | Express in form with given base |
| Difficulty | Moderate -0.8 This is a straightforward indices question requiring basic manipulation of powers (converting 8 to 2³) and simple algebraic rearrangement in part (a), followed by routine equation solving in part (b). The techniques are standard Core 1 material with no problem-solving insight needed, making it easier than average but not trivial since it requires multiple steps and careful index manipulation. |
| Spec | 1.02a Indices: laws of indices for rational exponents1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(8^x = 2^{3x}\) | B1 | Sight of \(8^x\) written as \(2^{3x}\), \((2^x)^3\) or \((2^3)^x\). May be implied. |
| \(\frac{8^x}{2^{x-1}} = 2^{3x-(x-1)}\) | M1 | Subtracts powers of 2. May be implied e.g. \(2^{2x\pm1}\). Do not award for incorrect application of laws of indices. |
| \(2^{2x+1} \Rightarrow a=2,\ b=1\) | A1 | \(a=2, b=1\) cao. Condone \(2^{2x+1}\). Do not award from incorrect work. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\sqrt{2} = 2^{1.5}\) | B1 | Sight or implied use of \(2^{1.5}\) or \(2^{\frac{3}{2}}\). Alternatively take logs and correctly apply power rule to achieve \((\text{"}2x+1\text{"})\ln 2 = \ln 2\sqrt{2}\) |
| \(\text{"}1.5\text{"} = \text{"}2x+1\text{"} \Rightarrow x = \ldots\) | M1 | Sets powers of 2 equal and solves for \(x\). Condone sign slips only. |
| \(x = \frac{1}{4}\) | A1 | cao |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $8^x = 2^{3x}$ | B1 | Sight of $8^x$ written as $2^{3x}$, $(2^x)^3$ or $(2^3)^x$. May be implied. |
| $\frac{8^x}{2^{x-1}} = 2^{3x-(x-1)}$ | M1 | Subtracts powers of 2. May be implied e.g. $2^{2x\pm1}$. Do not award for incorrect application of laws of indices. |
| $2^{2x+1} \Rightarrow a=2,\ b=1$ | A1 | $a=2, b=1$ cao. Condone $2^{2x+1}$. Do not award from incorrect work. |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\sqrt{2} = 2^{1.5}$ | B1 | Sight or implied use of $2^{1.5}$ or $2^{\frac{3}{2}}$. Alternatively take logs and correctly apply power rule to achieve $(\text{"}2x+1\text{"})\ln 2 = \ln 2\sqrt{2}$ |
| $\text{"}1.5\text{"} = \text{"}2x+1\text{"} \Rightarrow x = \ldots$ | M1 | Sets powers of 2 equal and solves for $x$. Condone sign slips only. |
| $x = \frac{1}{4}$ | A1 | cao |
---\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$ and the value of $b$ for which $\frac { 8 ^ { x } } { 2 ^ { x - 1 } } \equiv 2 ^ { a x + b }$
\item Hence solve the equation $\frac { 8 ^ { x } } { 2 ^ { x - 1 } } = 2 \sqrt { 2 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2019 Q2 [6]}}