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

Hint: To solve the given equation we will first use the property of log $\log \left( MN \right)=\log M+ \log N$ . Now if we use the definition of log we will get a quadratic equation in x. We will solve the quadratic equation by using the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Hence we will get the solution of the given equation.

Complete step-by-step solution:
Let us first understand the concept of logarithm.
Logarithm of a number x is the exponent to which base must be raised to produce the given number.
Hence let us say we have ${{10}^{2}}=100$
In logarithm form this can be written as ${{\log }_{10}}\left( 100 \right)=2$ in which 10 is called the base of logarithm.
Now if the base of logarithm is taken as e then the log is called natural logarithm and is written as ln.
Now let us understand three basic properties of logarithm.
${{\log }_{b}}(MN)={{\log }_{b}}\left( M \right)+{{\log }_{b}}\left( N \right)$
${{\log }_{b}}\left( \dfrac{M}{N} \right)={{\log }_{b}}\left( M \right)-{{\log }_{b}}\left( N \right)$
And ${{\log }_{b}}{{\left( M \right)}^{p}}=p{{\log }_{b}}\left( M \right)$
Now consider the given expression $\ln \left( x \right)+\ln \left( x-1 \right)=1$
Now we know that ${{\log }_{b}}(MN)={{\log }_{b}}\left( M \right)+{{\log }_{b}}\left( N \right)$ .
Hence using this we get,
$\Rightarrow \ln \left[ \left( x \right)\left( x-1 \right) \right]=1$
Now we know ln is nothing but the logarithm of base e. Hence using the definition of logarithm we get,
$\Rightarrow x\left( x-1 \right)={{e}^{1}}$
$\Rightarrow {{x}^{2}}-x-e=0$
Now this is a quadratic equation of the form $a{{x}^{2}}+bx+c=0$ where a = 1, b = - 1 and c = - e.
We know that the roots of the quadratic equation are given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Substituting the values of a, b and c in the formula we get,
$\begin{align}
  & \Rightarrow x=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 1 \right)\left( -e \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{1\pm \sqrt{1+4e}}{2} \\
\end{align}$
Hence the solution of the given equation $x=\dfrac{1\pm \sqrt{1+4e}}{2}$ .

Note: Note that logarithm with natural base is nothing but the inverse function of e. Hence ${{e}^{\ln \left( x \right)}}=x$ . Similarly we can say e is the inverse function of log and hence $\ln \left( {{e}^{x}} \right)=x$. Note that $\ln \left( e \right)=1$. Hence while solving logarithmic equations we can also use e function to eliminate log.