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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
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1 Answer 1

up vote 3 down vote accepted

As Thomas Andrews has pointed out, the inequalities in your third and fourth lines are the wrong way around. Having said that, I'm not sure whether your approach will work even if you make these corrections. Instead I present an alternative method which uses a little bit of differential calculus.

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.

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