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

Hint: We can clearly observe that the question given is in logarithmic form. We can solve this question by using some basic logarithmic formulae. And then some simplification is to be done to get the question solved.

Complete step by step answer:
From the question it had been given that $ \ln x+\ln \left( x+1 \right)=1 $
To get the above equation more simplified we have to use the basic formula of logarithms.
The formula to be used to solve the given question is shown below $ \ln a+\ln b=\ln ab $
By applying the above formula to the given question, we get $ \ln x+\ln \left( x+1 \right)=1 $
 $ \Rightarrow \ln \text{ x}\left( x+1 \right)=1 $
Hence, $ x\left( x+1 \right)=e $
On furthermore simplifying the above equation we get $ {{x}^{2}}+x-e=0 $
We can clearly observe that it is in the form of a quadratic equation.
We know that general form of quadratic equation is $ a{{x}^{2}}+bx+c=0 $
By comparing coefficients of both the equations, we get
 $ \begin{align}
  & a=1 \\
 & b=1 \\
 & c=-e \\
\end{align} $
We get the values of the coefficients, so we can solve the quadratic equation we got by using the quadratic formula.
Formula for quadratic formula is shown below: $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $
Substitute the values of coefficients we got in the quadratic formula.
By substituting the values of coefficients in the quadratic formula, we get
 $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $
 $ \Rightarrow x=\dfrac{-1\pm \sqrt{{{\left( 1 \right)}^{2}}-4.1.-e}}{2\left( 1 \right)} $
 $ \Rightarrow x=\dfrac{-1\pm \sqrt{1+4e}}{2} $
 $ \Rightarrow x=\dfrac{-1\pm 3.446}{2} $
From this we can say that $ x=\dfrac{2.446}{2} $ or $ x=-\dfrac{4.446}{2} $
After further simplifying we will have $ x=1.223 $ or $ x=-2.223 $

The logarithm of the negative value is invalid.
As negative value is not in our domain $ x=-2.223 $ is not a solution for the given question.

Therefore $ x=1.223 $ is the solution for the given logarithmic question.


Note:
We should be well aware of the logarithms. We should also be well aware of the basic formula and properties of logarithms. We should be very careful while doing the calculation of the quadratic formula. We should be well aware of the steps to be taken to solve the given question. Similarly, we can solve this type of other problems like $ \ln \left( x+1 \right)-\ln x=1\Rightarrow \ln \dfrac{x+1}{x}=1\Rightarrow \dfrac{x+1}{x}=e\Rightarrow 1+\dfrac{1}{x}=e\Rightarrow x=\dfrac{1}{e-1} $ .