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.