Question

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

Answer 1

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

& \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{aligned}$$ $$\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.