Edexcel C1 2016 June — Question 2 2 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Year2016
SessionJune
Marks2
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndices and Surds
TypeExpress in form with given base
DifficultyEasy -1.3 This is a straightforward index law question requiring only the recognition that 9 = 3² and application of power laws. It's a routine C1 exercise with minimal steps (rewrite base, multiply indices) and no problem-solving element, making it easier than average.
Spec1.02a Indices: laws of indices for rational exponents

Express \(9 ^ { 3 x + 1 }\) in the form \(3 ^ { y }\), giving \(y\) in the form \(a x + b\), where \(a\) and \(b\) are constants.
(2)

Question 2:
\[9^{3x+1} = 3^y\]
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(3^{2(3x+1)}\) or \((3^2)^{3x+1}\) or \((3^{(3x+1)})^2\) or \(3^{3x+1} \times 3^{3x+1}\) etc.; or \(y = 2(3x+1)\)M1 Expresses \(9^{3x+1}\) correctly as a power of 3, or expresses \(3^y\) correctly as a power of 9, or expresses \(y\) correctly in terms of \(x\). Not for just \(3^2 = 9\)
\(= 3^{6x+2}\) or \(y = 6x+2\) or \(a=6, b=2\)A1 cao (isw if necessary)
Special case: \(3^{6x+1}\) only scores M1A0
Alternative using logs:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\((3x+1)\log 9 = y\log 3\)M1 Use power law correctly on both sides
\(y = \frac{\log 9}{\log 3}(3x+1)\) then \(y = 6x+2\)A1 cao
Total: 2 marks
# Question 2:

$$9^{3x+1} = 3^y$$

| Answer/Working | Mark | Guidance |
|---|---|---|
| $3^{2(3x+1)}$ or $(3^2)^{3x+1}$ or $(3^{(3x+1)})^2$ or $3^{3x+1} \times 3^{3x+1}$ etc.; or $y = 2(3x+1)$ | M1 | Expresses $9^{3x+1}$ correctly as a power of 3, or expresses $3^y$ correctly as a power of 9, or expresses $y$ correctly in terms of $x$. **Not for just $3^2 = 9$** |
| $= 3^{6x+2}$ or $y = 6x+2$ or $a=6, b=2$ | A1 | cao (isw if necessary) |

**Special case:** $3^{6x+1}$ only scores M1A0

**Alternative using logs:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(3x+1)\log 9 = y\log 3$ | M1 | Use power law correctly on both sides |
| $y = \frac{\log 9}{\log 3}(3x+1)$ then $y = 6x+2$ | A1 | cao |

**Total: 2 marks**

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Express $9 ^ { 3 x + 1 }$ in the form $3 ^ { y }$, giving $y$ in the form $a x + b$, where $a$ and $b$ are constants.\\
(2)\\

\hfill \mbox{\textit{Edexcel C1 2016 Q2 [2]}}