Solving inequality with logarithms.

Question

I was playing around and found this

$$x\log a\le a-1$$

Solve for $x$ in the above equation, where $a>0$

My attempt

$$\log a\le \frac ax-\frac1x$$ $$a\le \frac{\exp(a/x)}{\exp(1/x)}$$

But I don't really see any way to proceed

I'm especially interested in knowing if some value of $x$ always solves the inequality.

Answer 1

This is to show that $x=1$ always works. So we need to show that $$f(a)=(a-1)-\log a \ge 0, \quad \forall a> 0$$

$$f'(a) = 1-\frac1a = \begin{cases} < 0, && a < 1 \\ > 0, && a> 1 \end{cases}$$

So the function is decreasing for $a < 1$ and increasing for $a> 1$, hence it achieves its minimum when $a=1$, i.e. $f(1)=0$. Hence $f(a) \ge 0$, so $x=1$ is always a solution.

Answer 2

Hints:

You should discuss that $\log a> 0$, $\log a<0$ and $\log a=0$ .

Answer 3

we have to distinguish two cases if $0<a<1$ then we have $x\geq \frac{a-1}{\log(a)}$ in the case that $a>1$ we have $x\le \frac{a-1}{\log(a)}$ and for $a=1$ we have $0\le 0$

Answer 4

To follow up on Macavity's excellent answer, here's a proof that $x = 1$ is the only value of $x$ that satisfies $x \log a \leq a - 1$ for all positive real values of $a.$

Let $x \log a \leq a - 1.$ For $a > 1,$ we have $\log a > 0,$ and therefore $$x \leq \frac{a - 1}{\log a}.$$ Now consider what happens for values of $a$ greater than but very close to $1.$ We have $$\lim_{a\to 1^+} \frac{a - 1}{\log a} = \lim_{a\to 1^+} \frac{1}{1/a} = 1.$$ If we set $x$ to any value greater than $1,$ we will then have $x > \frac{a - 1}{\log a}$ and therefore $x \log a > a - 1$ for some values of $a.$

For $a < 1,$ we have $\log a < 0,$ and therefore $$x \geq \frac{a - 1}{\log a}.$$ We have $$\lim_{a\to 1^+} \frac{a - 1}{\log a} = \lim_{a\to 1^+} \frac{1}{1/a} = 1.$$ So if $x < 1$ we will then have $x < \frac{a - 1}{\log a}$ and therefore $x \log a > a - 1$ for some values of $a.$ The value $x = 1$ is the only remaining value that satisfies $x \log a \leq a - 1$ for all positive real values of $a.$