1) Considering e =2.72
Then let's plug it in the inequality, and calculate the natural logarithm.
[tex]\begin{gathered} xe^x\ge7 \\ x2.72^x\ge7 \\ 2.72^x\ge\frac{7}{x}^{} \\ \ln 2.72^x\ge\ln (\frac{7}{x}) \\ x\text{ }\ge1.52 \end{gathered}[/tex]2) Then option 5 is the answer
X≥ 1.52
