How do you simplify $ {e^{1 + \ln x}} $ ?

Hint: In order to determine the value of the above question ,use logarithmic property $ {a^{m + n}} = \left( {{a^m}} \right)\left( {{a^n}} \right) $ to split the function into product and then use the fact that $ {e^{\log (n)}} = n $ to obtain the desired result

Complete step-by-step answer:
To solve the given question, we must know the properties of logarithms and with the help of them we are going to rewrite our question.
But we also need to know that the number $ e $ and $ \log n $ are actually inverses of each other.
First, we are going to rewrite the number with the help of the exponential identity that $ {a^{m + n}} = \left( {{a^m}} \right)\left( {{a^n}} \right) $
 $ \Rightarrow \left( {{e^1}} \right)\left( {{e^{\ln x}}} \right) $
Now using the fact that $ {e^{\log (n)}} = n $
 $
   \Rightarrow \left( e \right)\left( x \right) \\
   \Rightarrow ex \;
  $
Or
Putting the value of constant $ e = 2.71828 $
 $ \Rightarrow (2.71828)x $
Therefore, the simplification of $ {e^{1 + \ln x}} $ is equal to $ ex $ or $ (2.71828)x $ .
So, the correct answer is “ $ ex $ or $ (2.71828)x $ ”.

Note: We use the properties of logarithms to obtain results. Be aware about the base of log and proceed accordingly. we make use following formulae to solve the above question
Formula:
 $
  n\log m = \log {m^n} \\
 $
$ {e^{\log (n)}} = n $
 $ {a^{m + n}} = \left( {{a^m}} \right)\left( {{a^n}} \right) $