Question
Question: How do you simplify \( {e^{1 + \ln x}} \) ?...
How do you simplify e1+lnx ?
Solution
Hint : In order to determine the value of the above question ,use logarithmic property am+n=(am)(an) to split the function into product and then use the fact that elog(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 logn 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)
⇒(e1)(elnx)
Now using the fact that elog(n)=n
⇒(e)(x) ⇒ex
Or
Putting the value of constant e=2.71828
⇒(2.71828)x
Therefore, the simplification of e1+lnx 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:
nlogm=logmn
elog(n)=n
am+n=(am)(an)