# Find the smallest value of $a$ that a solution for this equation exists?

How to find the smallest value of $a$ that a solution for this equation exists?

$$2\ln x + \frac{5}{x} < a$$

That are the steps I've done, and have no idea what to do next:

\begin{align*} 2\ln x &< -\frac{5}{x} + a\\ \ln x^2 &< -\frac{5}{x} + a\\ e^{-\frac{5}{x} + a} &< x^2\\ e^{-5} \cdot e^{\frac{1}{x}} \cdot e^a &< x^2 \end{align*}

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$u<v$ if and only if $e^u<e^v$. You've instead concluded $e^v<e^u$. – Thomas Andrews Jan 23 '13 at 12:52

Let $f(x) = 2\ln x + \frac{5}{x}$ and note that it has domain $(0, \infty)$. The only way the inequality $f(x) < a$ can't have solutions is if $f(x) \geq a$ for all $x \in (0, \infty)$. In particular, if we know that $f$ has a global minimum (it need not have one), then for any $a$ less than or equal to that, the inequality will have no solutions.
You should be able to use calculus to determine whether $f$ has a global minimum, and if so, what it is.