Exponents & Logs

Both logarithms are defined only when $x – 1 > 0$ and $x + 1 > 0$, so $x > 1$. [1]

$log_{3}big((x-1)(x+1)big) = 2$, hence $log_{3}(x^{2} – 1) = 2$. [2]

$x^{2} – 1 = 3^{2} = 9$, so $x^{2} = 10$ and $x = pmsqrt{10}$. [2]

Only $x = sqrt{10}$ satisfies $x > 1$. [1]

Total: 6 marks