Given that \(u = \tanh x\), use the definitions of \(\sinh x\) and \(\cosh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\) to show that
$$x = \frac { 1 } { 2 } \ln \left( \frac { 1 + u } { 1 - u } \right)$$
Show that the equation
$$3 \operatorname { sech } ^ { 2 } x + 7 \tanh x = 5$$
can be written as
$$3 \tanh ^ { 2 } x - 7 \tanh x + 2 = 0$$
Show that the equation
$$3 \tanh ^ { 2 } x - 7 \tanh x + 2 = 0$$
has only one solution for \(x\).
Find this solution in the form \(\frac { 1 } { 2 } \ln a\), where \(a\) is an integer.