(a)     Find the solution of the equation

\[\ln {2^{4x - 1}} = \ln {8^{x + 5}} + {\log _2}{16^{1 - 2x}},\]

expressing your answer in terms of \(\ln 2\).

(b)     Using this value of x, find the value of a for which \({\log _a}x = 2\), giving your answer to three decimal places.

(a)     rewrite the equation as \((4x - 1)\ln 2 = (x + 5)\ln 8 + (1 - 2x){\log _2}16\)     (M1)

\((4x - 1)\ln 2 = (3x + 15)\ln 2 + 4 - 8x\)     (M1)(A1)

\(x = \frac{{4 + 16\ln 2}}{{8 + \ln 2}}\)     A1

(b)     \(x = {a^2}\)     (M1)

\(a = 1.318\)     A1

Note: Treat 1.32 as an AP.

Award A0 for ±.

[6 marks]

A more difficult question. Many candidates failed to read the question carefully so did not express x in terms of \(\ln 2\).