Solveeit Logo

Question

Question: How do you simplify \( {e^{1 + \ln x}} \) ?...

How do you simplify e1+lnx{e^{1 + \ln x}} ?

Explanation

Solution

Hint : In order to determine the value of the above question ,use logarithmic property am+n=(am)(an){a^{m + n}} = \left( {{a^m}} \right)\left( {{a^n}} \right) to split the function into product and then use the fact that elog(n)=n{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 ee and logn\log n are actually inverses of each other.
First, we are going to rewrite the number with the help of the exponential identity that am+n=(am)(an){a^{m + n}} = \left( {{a^m}} \right)\left( {{a^n}} \right)
(e1)(elnx)\Rightarrow \left( {{e^1}} \right)\left( {{e^{\ln x}}} \right)
Now using the fact that elog(n)=n{e^{\log (n)}} = n
(e)(x) ex   \Rightarrow \left( e \right)\left( x \right) \\\ \Rightarrow ex \;
Or
Putting the value of constant e=2.71828e = 2.71828
(2.71828)x\Rightarrow (2.71828)x
Therefore, the simplification of e1+lnx{e^{1 + \ln x}} is equal to exex or (2.71828)x(2.71828)x .
So, the correct answer is “ exex or (2.71828)x(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:
nlogm=logmn  n\log m = \log {m^n} \\\
elog(n)=n{e^{\log (n)}} = n
am+n=(am)(an){a^{m + n}} = \left( {{a^m}} \right)\left( {{a^n}} \right)