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

Hint: In the given question, we have been asked to solve the logarithmic expression for the value of ‘x’. In order to solve the question, first we need to simplify the equation by taking out common terms and then later we need exponentiation to both sides of the expression. Then simplifying the expression as exponents undo logarithmic functions. Later we completed the square of the equation and simplified further for the value of ‘x’. In this way we will get the required answer.

Complete step-by-step solution:
We have given that,
\[\ln x+\ln \left( x-1 \right)=1\]
Taking out the common term, we obtained
\[\Rightarrow \ln \left( x\left( x-1 \right) \right)=1\]
Simplifying the above logarithmic expression, we get
\[\Rightarrow \ln \left( {{x}^{2}}-x \right)=1\]
Taking exponential both the sides i.e. putting both sides to the power of ‘e’,
\[\Rightarrow {{e}^{\ln \left( {{x}^{2}}-x \right)}}={{e}^{1}}\]
Simplifying using the exponents and logarithm properties, we get
\[\Rightarrow {{x}^{2}}-x=e\]
Completing the square in the above equation, we obtained
\[\Rightarrow {{x}^{2}}-x+{{\left( \dfrac{1}{2} \right)}^{2}}=e+{{\left( \dfrac{1}{2} \right)}^{2}}\]
As we know that, \[{{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}\]
Therefore, applying the identity we obtained
\[\Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}=e+\dfrac{1}{4}\]
Simplifying the RHS by taking the LCM, we obtained
\[\Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}=\dfrac{4e+1}{4}\]
Transposing the power of 2 to the RHS, we get
\[\Rightarrow \left( x-\dfrac{1}{2} \right)=\sqrt{\dfrac{4e+1}{4}}=\pm \dfrac{\sqrt{4e+1}}{2}\]
Adding \[\dfrac{1}{2}\]to both the sides of the equation, we get
\[\Rightarrow x=\pm \dfrac{\sqrt{4e+1}}{2}+\dfrac{1}{2}=\dfrac{1\pm \sqrt{4e+1}}{2}\]
As we know that the value of natural log is always positive, thus eliminating the negative sign.
We obtained,
\[\Rightarrow x=\dfrac{1+\sqrt{4e+1}}{2}\]
Hence, it is the required answer.

Hence the correct answer is \[\Rightarrow x=\dfrac{1+\sqrt{4e+1}}{2}\]

Note: While solving these types of questions, students need to remember that exponential function undo the logarithmic function. In the given question, we need to find the value of ‘x’. To solve these types of questions, we used the basic formulas of logarithm. Students should always require to keep in mind all the formulae of logarithmic function for solving the question easily. After applying log formulae to the equation, we need to solve the equation in the way we solve general equations.