Solveeit Logo

Question

Question: How do you expand \(\ln \left( x{{e}^{x}} \right)\) ?...

How do you expand ln(xex)\ln \left( x{{e}^{x}} \right) ?

Explanation

Solution

In this question, we have to expand the given logarithmic expression. Thus, we will use the logarithmic formula and the log-exponential formula to get the solution. Now, we know that log function is the inverse of the exponential function. Thus, in this problem, we will first, apply the logarithmic formula log(ab)=loga+logb\log \left( ab \right)=\log a+\log b in the given expression, where a=xa=x and b=exb={{e}^{x}} . After that, we will again apply the logarithmic-exponential formula logea=a\log {{e}^{a}}=a in the expression, to get the required solution for the problem.

Complete step by step solution:
According to the problem, we have to expand the given logarithmic expression.
Thus, we will use the logarithmic and the log-exponential formula to get the solution.
The logarithmic expression given to us ln(xex)\ln \left( x{{e}^{x}} \right) -------- (1)
Now, we will first apply the logarithmic formula log(ab)=loga+logb\log \left( ab \right)=\log a+\log b in equation (1) where a=xa=x and b=exb={{e}^{x}} , thus we get
logx+log(ex)\Rightarrow \log x+\log \left( {{e}^{x}} \right)
In the last, we will again apply the logarithmic formula logea=a\log {{e}^{a}}=a in the above expression, we get
logx+x\Rightarrow \log x+x which is the required solution
Therefore, for the logarithmic expression ln(xex)\ln \left( x{{e}^{x}} \right) , thus its expanded value is equal to logx+x\log x+x.

Note: While solving this problem, do the step by step calculations properly to avoid confusion and mathematical error. Do not forget to mention the formula you are using to get the accurate answer. Also remember that log function is the inverse of the exponential function, which implies that when we have the log and the exponential function together, we get the value equal to the power of the exponential function, that is the formula for the same is logea=a\log {{e}^{a}}=a.