June 2019 Paper 2 Q1

EdexcelCurrent spec3 marksLogs & Exponentials Quadratics

1. Given

\[2^x \times 4^y = \frac{1}{2\sqrt{2}}\]

express \(y\) as a function of \(x\). (3)

Mark scheme

Scheme	Marks	AO
\(2^x \times 4^y = \dfrac{1}{2\sqrt{2}}\ \left\{= \dfrac{\sqrt{2}}{4}\right\}\)
Way 1 \(2^x \times 2^{2y} = 2^{-\frac{3}{2}}\)	B1	1.1b
\(2^{x+2y} = 2^{-\frac{3}{2}} \Rightarrow x + 2y = -\dfrac{3}{2} \Rightarrow y = \ldots\)	M1	2.1
E.g. \(y = -\dfrac{1}{2}x - \dfrac{3}{4}\) or \(y = -\dfrac{1}{4}(2x + 3)\)	A1	1.1b
	(3)
	(3 marks)

Notes

Special Case

If 0 marks are scored on application of the mark scheme then allow Special Case B1 M0 A0 (total of 1 mark) for any of

\(2^x \times 4^y \to 2^{x+2y}\)
\(2^x \times 4^y \to 4^{\frac{1}{2}x + y}\)
\(\dfrac{1}{2^x 2\sqrt{2}} \to 2^{-x-\frac{3}{2}}\)
\(\log 2^x + \log 4^y \to x\log 2 + y\log 4\) or \(x\log 2 + 2y\log 2\)
\(\ln 2^x + \ln 4^y \to x\ln 2 + y\ln 4\) or \(x\ln 2 + 2y\ln 2\)
\(y = \log\left(\dfrac{1}{2^x 2\sqrt{2}}\right)\) o.e. {base of 4 omitted}

Way 2

Scheme	Marks	AO
\(\log(2^x \times 4^y) = \log\left(\dfrac{1}{2\sqrt{2}}\right)\)	B1	1.1b
\(\log 2^x + \log 4^y = \log\left(\dfrac{1}{2\sqrt{2}}\right)\) \(\Rightarrow x\log 2 + y\log 4 = \log 1 - \log(2\sqrt{2}) \Rightarrow y = \ldots\)	M1	2.1
\(y = \dfrac{-\log(2\sqrt{2}) - x\log 2}{\log 4}\ \left\{\Rightarrow y = -\dfrac{1}{2}x - \dfrac{3}{4}\right\}\)	A1	1.1b
	(3)

Way 3

Scheme	Marks	AO
\(\log(2^x \times 4^y) = \log\left(\dfrac{1}{2\sqrt{2}}\right)\)	B1	1.1b
\(\log 2^x + \log 4^y = \log\left(\dfrac{1}{2\sqrt{2}}\right) \Rightarrow \log 2^x + y\log 4 = \log\left(\dfrac{1}{2\sqrt{2}}\right) \Rightarrow y = \ldots\)	M1	2.1
\(y = \dfrac{\log\left(\dfrac{1}{2\sqrt{2}}\right) - \log(2^x)}{\log 4}\ \left\{\Rightarrow y = -\dfrac{1}{2}x - \dfrac{3}{4}\right\}\)	A1	1.1b
	(3)

Way 4

Scheme	Marks	AO
\(\log_2(2^x \times 4^y) = \log_2\left(\dfrac{1}{2\sqrt{2}}\right)\)	B1	1.1b
\(\log_2 2^x + \log_2 4^y = \log_2\left(\dfrac{1}{2\sqrt{2}}\right) \Rightarrow x + 2y = -\dfrac{3}{2} \Rightarrow y = \ldots\)	M1	2.1
E.g. \(y = -\dfrac{1}{2}x - \dfrac{3}{4}\) or \(y = -\dfrac{1}{4}(2x + 3)\)	A1	1.1b
	(3)

Way 5

Scheme	Marks	AO
\(4^{\frac{1}{2}x} \times 4^y = 4^{-\frac{3}{4}}\)	B1	1.1b
\(4^{\frac{1}{2}x + y} = 4^{-\frac{3}{4}} \Rightarrow \dfrac{1}{2}x + y = -\dfrac{3}{4} \Rightarrow y = \ldots\)	M1	2.1
E.g. \(y = -\dfrac{1}{2}x - \dfrac{3}{4}\) or \(y = -\dfrac{1}{4}(2x + 3)\)	A1	1.1b
	(3)

Notes for Question 1

Way 1

B1: Writes a correct equation in powers of 2 only

M1: Complete process of writing a correct equation in powers of 2 only and using correct index laws to obtain \(y\) written as a function of \(x\).

A1: \(y = -\dfrac{1}{2}x - \dfrac{3}{4}\) o.e.

Way 2, Way 3 and Way 4

B1: Writes a correct equation involving logarithms

M1: Complete process of writing a correct equation involving logarithms and using correct log laws to obtain \(y\) written as a function of \(x\).

A1: \(y = \dfrac{-\log(2\sqrt{2}) - x\log 2}{\log 4}\) or \(y = \dfrac{-\ln(2\sqrt{2}) - x\ln 2}{\ln 4}\) or \(y = \dfrac{\log\left(\dfrac{1}{2\sqrt{2}}\right) - \log(2^x)}{\log 4}\)
or \(y = -\dfrac{1}{2}x - \dfrac{3}{4}\) or \(y = -\dfrac{1}{4}(2x + 3)\) o.e.

Way 5

B1: Writes a correct equation in powers of 4 only

M1: Complete process of writing a correct equation in powers of 4 only and using correct index laws to obtain \(y\) written as a function of \(x\).

A1: \(y = -\dfrac{1}{2}x - \dfrac{3}{4}\) o.e.

Note: Allow equivalent results for A1 where \(y\) is written as a function of \(x\)

Note: You can ignore subsequent working following on from a correct answer.

Note: Allow B1 for \(2^x \times 4^y = \dfrac{1}{2\sqrt{2}} \Rightarrow 4^y = \dfrac{1}{2^x 2\sqrt{2}} \Rightarrow \log_4(4^y) = \log_4\left(\dfrac{1}{2^x 2\sqrt{2}}\right)\)
followed by M1 A1 for \(y = \log_4\left(\dfrac{1}{2^x 2\sqrt{2}}\right)\) or \(y = \log_4\left(\dfrac{2^{-x}}{2\sqrt{2}}\right)\) or \(y = \log_4\left(\dfrac{\sqrt{2}}{4(2^x)}\right)\)
or \(y = -\log_4\left(2^{x + \frac{3}{2}}\right)\) or \(y = -\log_4(\sqrt{2}(2^{x+1}))\)