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

How do you solve ln(x+1)ln(x2)=lnx\ln \left( x+1 \right)-\ln \left( x-2 \right)=\ln x?

Explanation

Solution

Try to simplify the equation by applying lnmlnn=lnmn\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=b±D2ax=\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(x+1)ln(x2)=lnx\ln \left( x+1 \right)-\ln \left( x-2 \right)=\ln x
As we know, lnmlnn=lnmn\ln m-\ln n=\ln \dfrac{m}{n}
ln(x+1x2)=lnx\Rightarrow \ln \left( \dfrac{x+1}{x-2} \right)=\ln x
Again since we know, elnm=m{{e}^{\ln m}}=m
Hence taking ‘e’ to the power both the sides, we get
eln(x+1x2)=elnx x+1x2=x \begin{aligned} & \Rightarrow {{e}^{\ln \left( \dfrac{x+1}{x-2} \right)}}={{e}^{\ln x}} \\\ & \Rightarrow \dfrac{x+1}{x-2}=x \\\ \end{aligned}
Cross multiplying, we get
x+1=x(x2) x+1=x22x x22xx1=0 x23x1=0 \begin{aligned} & \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{aligned}
Quadratic equation: In order to find the roots of a quadratic equation ax2+bx+c=0a{{x}^{2}}+bx+c=0, first we have to find it’s discriminant(‘D’) which is given by D=b24acD=\sqrt{{{b}^{2}}-4ac}. Then the roots of the quadratic equation will be x=b±D2ax=\dfrac{-b\pm \sqrt{D}}{2a}.
Now, let’s consider our quadratic equation x23x1=0{{x}^{2}}-3x-1=0
Here D=b24ac=(3)241(1)=9(4)=9+4=13D=\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=b±D2a=(3)±1321=3±132x=\dfrac{-b\pm \sqrt{D}}{2a}=\dfrac{-\left( -3 \right)\pm \sqrt{13}}{2\cdot 1}=\dfrac{3\pm \sqrt{13}}{2}
Either x=3132x=\dfrac{3-\sqrt{13}}{2} or x=3+132x=\dfrac{3+\sqrt{13}}{2}
For x=3132x=\dfrac{3-\sqrt{13}}{2}, the value of x is negative. So, lnx\ln xwill be undefined.
Hence, x=3+132x=\dfrac{3+\sqrt{13}}{2}is the solution of our equation.

Note:
For finding roots of the quadratic equation part, the formula x=b±b24ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} can be used directly in a single step instead of x=b±D2ax=\dfrac{-b\pm \sqrt{D}}{2a} (where D=b24acD=\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=3132x=\dfrac{3-\sqrt{13}}{2} can’t be considered because it gives a negative value as x=3132=33.6062=0.6062=0.303x=\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.