Show that a positive constant \(t\) satisfies \[ e^x>x^t \qquad\text{for every }x>0 \] if and only if \[ t<e. \]
Solution
For \(x>1\), define \[ h(x)\coloneqq\frac{x}{\log x}. \]
1. The function \(h\) has minimum value \(e\) on \((1,\infty)\), attained at\(x=e\).
Differentiation gives \[ h'(x) = \frac{\log x-1}{(\log x)^2}. \] Thus \(h'(x)<0\) for \(1<x<e\), \(h'(e)=0\), and \(h'(x)>0\) for \(x>e\). Therefore \(h\) decreases on \((1,e]\) and increases on \([e,\infty)\), so \[ h(x)\geq h(e)=e \] for every \(x>1\).
2. If \(0<t<e\), then\[ e^x>x^t \]for every \(x>0\).
If \(0<x\leq1\), then \[ x^t\leq1<e^x. \] If \(x>1\), then step 1 gives \[ t<e\leq\frac{x}{\log x}. \] Since \(\log x>0\), this implies \[ t\log x<x. \] Exponentiating gives \[ x^t<e^x. \]
3. If \(t\geq e\), then the inequality fails for \(x=e\).
At \(x=e\), \[ x^t=e^t\geq e^e=e^x. \] Thus the required strict inequality does not hold for every positive \(x\).
4. A positive constant \(t\) satisfies the required inequality for every\(x>0\) if and only if\[ \boxed{t<e}. \]
Step 2 proves sufficiency, and step 3 proves necessity.
5. Q.E.D.
Step 4 is exactly the desired equivalence.