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 $$b/a \ge 100$$ $$1/100 \ge b/a \ > 0$$

$\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
$a/b \le e^{-2}$.
