Solve the following system of equations.

\[{\log _{x + 1}}y = 2\]\[{\log _{y + 1}}x = \frac{1}{4}\]

\({\log _{x + 1}}y = 2\)

\({\log _{y + 1}}x = \frac{1}{4}\)

so \({\left( {x + 1} \right)^2} = y\)     A1

\({\left( {y + 1} \right)^{\frac{1}{4}}} = x\)     A1

EITHER

\({x^4} - 1 = {\left( {x + 1} \right)^2}\)     M1

\(x = - 1\), not possible     R1

\(x = 1.70\), \(y = 7.27\)     A1A1

OR

1
\({\left( {{x^2} + 2x + 2} \right)^{\frac{1}{4}}} - x = 0\)     M1

attempt to solve or graph of LHS     M1

\(x =1.70\), \(y = 7.27\)     A1A1

[6 marks]

This question was well answered by a significant number of candidates. There was evidence of good understanding of logarithms. The algebra required to solve the problem did not intimidate candidates and the vast majority noticed the necessity of technology to solve the final equation. Not all candidates recognized the extraneous solution and there were situations where a rounded value of \(x\) was used to calculate the value of \(y\) leading to an incorrect solution.