Exercise 3.4.2 solution

  1. \(3^{1 - x} = 3^2\). So \(1 - x = 2\). Thus \(x = -1\).
  2. \(\log_2( x^2 - x ) = \log_2 ( 2 )\). So \(x^2 - x = 2\). That is, \(x^2 - x - 2 = 0\). So \(x = 2\) or \(x = -1\). Since we can only take logarithms of positive numbers, we must check that \(x^2-x>0\) for both solutions before we accept them. We see that \(x^2-x>0\) both for \(x=2\) and for \(x=-1\), so we accept both solutions as valid.