Solve the following equations:

(a)     ${\log _2}(x - 2) = {\log _4}({x^2} - 6x + 12)$;

(b)     ${x^{\ln x}} = {{\text{e}}^{{{(\ln x)}^3}}}$.

(a)     ${\log _2}(x - 2) = {\log _4}({x^2} - 6x + 12)$

EITHER

${\log _2}(x - 2) = \frac{{{{\log }_2}({x^2} - 6x + 12)}}{{{{\log }_2}4}}$     M1

$2{\log _2}(x - 2) = {\log _2}({x^2} - 6x + 12)$

OR

$\frac{{{{\log }_4}(x - 2)}}{{{{\log }_4}2}} = {\log _4}({x^2} - 6x + 12)$     M1

$2{\log _4}(x - 2) = {\log _4}({x^2} - 6x + 12)$

THEN

${(x - 2)^2} = {x^2} - 6x + 12$     A1

${x^2} - 4x + 4 = {x^2} - 6x + 12$

$x = 4$     A1     N1

[3 marks]

(b)     ${x^{\ln x}} = {{\text{e}}^{{{(\ln x)}^3}}}$

taking ln of both sides or writing $x = {{\text{e}}^{\ln x}}$     M1

${(\ln x)^2} = {(\ln x)^3}$     A1

${(\ln x)^2}(\ln x - 1) = 0$     (A1)

$x = 1,{\text{ }}x = {\text{e}}$     A1A1     N2

Note:     Award second (A1) only if factorisation seen or if two correct

solutions are seen.

[5 marks]

Total [8 marks]

Part a) was answered well, and a very large proportion of candidates displayed familiarity and confidence with this type of change-of base equation.

In part b), good candidates were able to solve this proficiently. A number obtained only one solution, either through observation or mistakenly cancelling a $\ln x$ term. An incorrect solution $x = {{\text{e}}^3}$ was somewhat prevalent amongst the weaker candidates.