EXN.1.SL.TZ0.2

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

 

METHOD 1

$2 \ln x- \ln 9=4$

uses $m \ln x= \ln x^m$       (M1)

$\ln x^2- \ln 9=4$

uses $\ln a- \ln b=\ln \frac{a}{b}$       (M1)

$\ln \frac{x^2}{9}=4$

$\frac{x^2}{9}={\mathrm{e}}^4$       A1

$x^2=9{\mathrm{e}}^4\Rightarrow x=\sqrt{9{\mathrm{e}}^4} \left(x>0\right)$       A1

$x=3{\mathrm{e}}^2 \left(p=3, q=2\right)$       A1

 

METHOD 2

expresses $4$ as $4 \ln \mathrm{e}$ and uses $\ln x^m=m \ln x$       (M1)

$2 \ln x=2 \ln 3+4 \ln \mathrm{e} \left(\ln x=\ln 3+2 \ln \mathrm{e}\right)$       A1

uses $2 \ln \mathrm{e}=\ln {\mathrm{e}}^2$ and $\ln a+\ln b=\ln ab$       (M1)

$\ln x=\ln \left(3{\mathrm{e}}^2\right)$       A1

$x=3{\mathrm{e}}^2 \left(p=3, q=2\right)$       A1

 

METHOD 3

expresses $4$ as $4 \ln \mathrm{e}$ and uses $m \ln x=\ln x^m$       (M1)

$\ln x^2=\ln 3^2+\ln {\mathrm{e}}^4$       A1

uses $\ln a+\ln b=\ln ab$       (M1)

$\ln x^2=\ln \left(3^2 {\mathrm{e}}^4\right)$

$x^2=3^2 {\mathrm{e}}^4\Rightarrow x=\sqrt{3^2 {\mathrm{e}}^4} \left(x>0\right)$       A1

so $x=3{\mathrm{e}}^2 \left(x>0\right) \left(p=3, q=2\right)$       A1

 

[5 marks]

[N/A]