Solve ${\left( {{\text{ln}}\,x} \right)^2} - \left( {{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x} \right) < 2{\left( {{\text{ln}}\,2} \right)^2}$.

${\left( {{\text{ln}}\,x} \right)^2} - \left( {{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x} \right) - 2{\left( {{\text{ln}}\,2} \right)^2}\left( { = 0} \right)$

EITHER

${\text{ln}}\,x = \frac{{{\text{ln}}\,2 \pm \sqrt {{{\left( {{\text{ln}}\,2} \right)}^2} + 8{{\left( {{\text{ln}}\,2} \right)}^2}} }}{2}$     M1

$= \frac{{{\text{ln}}\,2 \pm 3\,{\text{ln}}\,2}}{2}$     A1

OR

$\left( {{\text{ln}}\,x - 2\,{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x + 2\,{\text{ln}}\,2} \right)\left( { = 0} \right)$     M1A1

THEN

${\text{ln}}\,x = 2\,{\text{ln}}\,2$ or $- {\text{ln}}\,2$     A1

$\Rightarrow x = 4$ or $x = \frac{1}{2}$       (M1)A1   

Note: (M1) is for an appropriate use of a log law in either case, dependent on the previous M1 being awarded, A1 for both correct answers.

solution is $\frac{1}{2} < x < 4$     A1

[6 marks]