How to find the range of unknowns from a logarithmic equation? If the equation $\log(ax)\cdot \log(bx) + 1 = 0$ with $a>0, b>0$ constants has a solution $x>0$, it follows that $$b/a \geq (?)$$ or $$(?)\geq b/a >(?)$$
Hints maybe (for both :p)?
EDIT: 
Answer $$b/a \ge 100$$ $$1/100 \ge b/a \ > 0$$
 A: I guess your question is related to solving the equation of $x$, and then you might be interested about some features/conditions about $x$. As I see it, if you assume for only real numbers, $x$ would always be positive if the solution exists.
To see this:
$$\log (ax)\cdot\log(bx) + 1 = 0$$ 
$$\log(ax)\cdot\bigg(\log(ax\cdot\frac{b}{a})\bigg) + 1 =0$$
$$\log(ax)\cdot\bigg(\log(ax) + \log(b) - \log(a)\bigg) + 1 =0$$
$$\log^2(ax) + \bigg(\log(b) - \log(a)\bigg)\cdot \log(ax) + 1 =0$$
If we let $y=\log(ax)$, then we get
$$y^2 +(\log(b) - \log(a))y + 1=0$$
which is a quadratic equation, whose solution is $\log(ax)=y=\frac{-(\log(a)-\log(b))\pm \sqrt{\Delta}}{2\times1}$, where $\Delta=(\log(b)-\log(a))^2-4\times1\times1.$
And you should be able to solve $\log(ax)$ if your $\Delta \ge 0$, and then you could get $x$. But as long as you get the solution, $x$ is garanteed to be positive, because you'll eventually have $\log(ax) = m$, and then $x = \frac{1}{a}e^m > 0$
But maybe you are interested in the condition that we have a solution for $x$, if that is case, we just need to make sure that $\Delta = (\log(b) - \log(a))^2 -4\ge 0$
That is $\log\frac{b}{a} \ge 2$ or $\log\frac{b}{a} \le -2$ for the solutions to exist; which means $\frac{b}{a} \ge e^2$ or $\frac{b}{a} \le \frac{1}{e^2}$.
A: $\begin{array}\\
-1
&=\log(ax)\cdot \log(bx) \\
&=(\log(a)+\log(x))(\log(b)+\log(x))\\
&=(A+X)(B+X)
\qquad\text{with } A=\log a, B=\log b, X = \log x\\
&=AB+X(A+B)+X^2\\
\end{array}
$
so
$X^2+X(A+B)+AB+1 = 0$.
Solving,
$\begin{array}\\
X
&=\dfrac{-A-B\pm\sqrt{A^2+2AB+B^2-4AB-4}}{2}\\
&=\dfrac{-A-B\pm\sqrt{A^2-2AB+B^2-4}}{2}\\
&=\dfrac{-A-B\pm\sqrt{(A-B)^2-4}}{2}\\
\end{array}
$
If this has a real solution,
then
$(A-B)^2 \ge 4$
so
$|A-B| \ge 2$
or
$2
\le |A-B|
= |\log(a)-\log(b)|
= |\log(a/b)|
$.
Therefore
$\log(a/b) \ge 2$
or
$\log(a/b) \le -2$.
In the first case,
$a/b \ge e^2$
(assuming natural logs);
in the second
$a/b \le e^{-2}$.
