C3 June 2007 Q1

EdexcelOld spec6 marksLogs & Exponentials

1. Find the exact solutions to the equations

(a) \(\ln x + \ln 3 = \ln 6\), (2)

(b) \(\mathrm{e}^x + 3\mathrm{e}^{-x} = 4\). (4)

Mark scheme (a)

Scheme	Marks
\(\ln 3x = \ln 6\) or \(\ln x = \ln\left(\dfrac{6}{3}\right)\) [implied by 0.69…] or \(\ln\left(\dfrac{3x}{6}\right) = 0\)	M1
\(x = 2\) (only this answer)	A1 (cso)
	(2)

Notes

Answer \(x = 2\) with no working or no incorrect working seen: M1A1

Beware \(x = 2\) from \(\ln x = \dfrac{\ln 6}{\ln 3} = \ln 2\) M0A0

\(\ln x = \ln 6 - \ln 3 \Rightarrow x = e^{(\ln 6\, -\, \ln 3)}\) allow M1, \(x = 2\) (no wrong working) A1

Mark scheme (b)

Scheme	Marks
\((\mathrm{e}^x)^2 - 4\mathrm{e}^x + 3 = 0\) (any 3 term form)	M1
\((\mathrm{e}^x - 3)(\mathrm{e}^x - 1) = 0\)
\(\mathrm{e}^x = 3\) or \(\mathrm{e}^x = 1\) Solving quadratic	M1 dep
\(x = \ln 3\), \(x = 0\) (or \(\ln 1\))	M1 A1
	(4)
	(6 marks)

Notes

1^st M1 for attempting to multiply through by \(\mathrm{e}^x\): Allow \(y\), \(X\), even \(x\), for \(\mathrm{e}^x\)
Be generous for M1 e.g. \(e^{2x} + 3 = 4\), \(e^{x^2} + 3 = 4e^x\), \(3y^2 + 1 = 12y\) (from \(3\mathrm{e}^{-x} = \dfrac{1}{3e^x}\)), \(\mathrm{e}^x + 3 = 4\mathrm{e}^x\)

2^nd M1 is for solving quadratic (may be by formula or completing the square) as far as getting two values for \(\mathrm{e}^x\) or \(y\) or \(X\) etc

3^rd M1 is for converting their answer(s) of the form \(\mathrm{e}^x = k\) to \(x = \ln k\) (must be exact)
A1 is for \(\ln 3\) and \(\ln 1\) or 0 (Both required and no further solutions)