Question
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(x−2)=lnx?
Solution
Try to simplify the equation by applying lnm−lnn=lnnm first, then taking ‘e’ to the power both the sides. After getting the quadratic equation apply x=2a−b±Dto 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(x−2)=lnx
As we know, lnm−lnn=lnnm
⇒ln(x−2x+1)=lnx
Again since we know, elnm=m
Hence taking ‘e’ to the power both the sides, we get
⇒eln(x−2x+1)=elnx⇒x−2x+1=x
Cross multiplying, we get
⇒x+1=x(x−2)⇒x+1=x2−2x⇒x2−2x−x−1=0⇒x2−3x−1=0
Quadratic equation: In order to find the roots of a quadratic equation ax2+bx+c=0, first we have to find it’s discriminant(‘D’) which is given by D=b2−4ac. Then the roots of the quadratic equation will be x=2a−b±D.
Now, let’s consider our quadratic equation x2−3x−1=0
Here D=b2−4ac=(−3)2−4⋅1⋅(−1)=9−(−4)=9+4=13
So, x=2a−b±D=2⋅1−(−3)±13=23±13
Either x=23−13 or x=23+13
For x=23−13, the value of x is negative. So, lnxwill be undefined.
Hence, x=23+13is the solution of our equation.
Note:
For finding roots of the quadratic equation part, the formula x=2a−b±b2−4ac can be used directly in a single step instead of x=2a−b±D (where D=b2−4ac ). This is to avoid the calculation error for the longer calculations. Among the two values of ‘x’ that we got the value x=23−13 can’t be considered because it gives a negative value as x=23−13=23−3.606=−20.606=−0.303 and the logarithm of a negative value is always undefined. So, ultimately the other is the solution which will satisfy the equation.