Question

How do you solve $\ln \left( x+1 \right)-\ln \left( x-2 \right)=\ln x$?

Answer 1

Hint: Try to simplify the equation by applying $\ln m-\ln n=\ln \dfrac{m}{n}$ first, then taking ‘e’ to the power both the sides. After getting the quadratic equation apply $x=\dfrac{-b\pm \sqrt{D}}{2a}$to find the roots of the quadratic equation. Among the two roots decide the solution by considering all the criteria of the question.

Complete step by step answer:
Solving an equation means, we have to find the value of ‘x’ for which the equation gets satisfied.
Considering our equation $\ln \left( x+1 \right)-\ln \left( x-2 \right)=\ln x$
As we know, $\ln m-\ln n=\ln \dfrac{m}{n}$
$\Rightarrow \ln \left( \dfrac{x+1}{x-2} \right)=\ln x$
Again since we know, ${{e}^{\ln m}}=m$
Hence taking ‘e’ to the power both the sides, we get
$\begin{align}
  & \Rightarrow {{e}^{\ln \left( \dfrac{x+1}{x-2} \right)}}={{e}^{\ln x}} \\
 & \Rightarrow \dfrac{x+1}{x-2}=x \\
\end{align}$
Cross multiplying, we get
$\begin{align}
  & \Rightarrow x+1=x\left( x-2 \right) \\
 & \Rightarrow x+1={{x}^{2}}-2x \\
 & \Rightarrow {{x}^{2}}-2x-x-1=0 \\
 & \Rightarrow {{x}^{2}}-3x-1=0 \\
\end{align}$
Quadratic equation: In order to find the roots of a quadratic equation $a{{x}^{2}}+bx+c=0$, first we have to find it’s discriminant(‘D’) which is given by $D=\sqrt{{{b}^{2}}-4ac}$. Then the roots of the quadratic equation will be $x=\dfrac{-b\pm \sqrt{D}}{2a}$.
Now, let’s consider our quadratic equation ${{x}^{2}}-3x-1=0$
Here $D=\sqrt{{{b}^{2}}-4ac}=\sqrt{{{\left( -3 \right)}^{2}}-4\cdot 1\cdot \left( -1 \right)}=\sqrt{9-\left( -4 \right)=}\sqrt{9+4}=\sqrt{13}$
So, $x=\dfrac{-b\pm \sqrt{D}}{2a}=\dfrac{-\left( -3 \right)\pm \sqrt{13}}{2\cdot 1}=\dfrac{3\pm \sqrt{13}}{2}$
Either $x=\dfrac{3-\sqrt{13}}{2}$ or $x=\dfrac{3+\sqrt{13}}{2}$
For $x=\dfrac{3-\sqrt{13}}{2}$, the value of x is negative. So, $\ln x$will be undefined.
Hence, $x=\dfrac{3+\sqrt{13}}{2}$is the solution of our equation.

Note:
For finding roots of the quadratic equation part, the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ can be used directly in a single step instead of $x=\dfrac{-b\pm \sqrt{D}}{2a}$ (where $D=\sqrt{{{b}^{2}}-4ac}$ ). This is to avoid the calculation error for the longer calculations. Among the two values of ‘x’ that we got the value $x=\dfrac{3-\sqrt{13}}{2}$ can’t be considered because it gives a negative value as $x=\dfrac{3-\sqrt{13}}{2}=\dfrac{3-3.606}{2}=-\dfrac{0.606}{2}=-0.303$ and the logarithm of a negative value is always undefined. So, ultimately the other is the solution which will satisfy the equation.