How do you solve \[{{\left( \ln (x) \right)}^{2}}+\ln (x)-6=0\]?

Hint: To solve the given question, we should know how to find the roots of a quadratic equation. For a given quadratic equation \[a{{x}^{2}}+bx+c=0\], using the formula method, we can find the roots of the equation as \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\], by substituting the coefficients in the formula we can find the roots of the equation. Also, we should know that, \[\ln a=b\Rightarrow {{e}^{b}}=a\].

Complete step-by-step solution:
The given equation is \[{{\left( \ln (x) \right)}^{2}}+\ln (x)-6=0\]. By substituting \[\ln (x)=t\] in this equation, it can be expressed as \[{{t}^{2}}+t-6=0\]. This is a quadratic equation in t. we know that for a given quadratic equation \[a{{x}^{2}}+bx+c=0\], using the formula method, we can find the roots of the equation as \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
Substituting the values of the coefficients in the above formula, we get
\[\begin{align}
  & \Rightarrow t=\dfrac{-1\pm \sqrt{{{1}^{2}}-4(1)(-6)}}{2(1)} \\
 & \Rightarrow t=\dfrac{-1\pm \sqrt{25}}{2} \\
 & \Rightarrow t=\dfrac{-1\pm 5}{2} \\
\end{align}\]
\[\Rightarrow t=\dfrac{-1+5}{2}\] or \[t=\dfrac{-1-5}{2}\]
\[\Rightarrow t=\dfrac{4}{2}=2\] or \[t=\dfrac{-6}{2}=-3\]
Hence, the roots of the equation are \[t=2\] and \[t=-3\].

Using the substitution, we can say that \[\ln x=2\] or \[\ln x=-3\].
Using the property, \[\ln a=b\Rightarrow {{e}^{b}}=a\]. For the above two values, we get
For \[\ln x=2\], we get \[x={{e}^{2}}\].
For\[\ln x=-3\], we get \[x={{e}^{-3}}\].
Hence, the solution values for the given equation are \[x={{e}^{2}}\] or \[x={{e}^{-3}}\].

Note: We can check whether the solution is correct or not by substituting the values in the given equation.
Substituting \[x={{e}^{2}}\] in the equation, we get
\[{{\left( \ln ({{e}^{2}}) \right)}^{2}}+\ln ({{e}^{2}})-6=0\]
Using the property of logarithm \[\ln {{a}^{b}}=b\ln a\], the above expression can be expressed as
\[\Rightarrow {{\left( 2 \right)}^{2}}+2-6=0\]
Simplifying the above equation, we get
\[\Rightarrow 0=0\]
As the above statement is true, we can say that \[x={{e}^{2}}\] is the solution of the given equation.
Similarly, we can also check for \[x={{e}^{-3}}\], by substituting its value in the equation.