# If $\log(ax)\log(bx) +1=0$ has a solution $x>0$, then find bounds on $b/a$

If equation $$\log(ax)\log(bx) +1=0$$ with constants $$\;a>0,\; b>0\;$$ has a solution $$x>0$$, it follows that $$\frac{b}{a} \ge ???$$ or $$???\ge\frac{b}{a}\gt???$$

Fill all in the blank.

To be honest, I am very lost here and not sure how I can get into $$\frac{b}{a}$$ part. The answers provided were $$100, 1/100,\;$$ and $$\;0\;$$ respectively.

I would like to hear the perspective of how other people think about this problem. Looking forward to hearing from you!

• The expression is symmetric in $a,b$. Whatever inequality hold for $\frac ba$ must also hold for $\frac ab$. – lulu Feb 4 at 16:36

The equation can be rewritten as $$(\log a + \log x))(\log b+ \log x) +1=0.$$ Set $$t=\log x$$ and expand, the obtained quadratic equation $$t^2+t(\log a + \log b) + \log a \cdot \log b +1=0.$$ The discriminant $$D=(\log {a\over b}-2)(\log {a\over b}+2)$$ must satisfy $$(\log {a\over b}-2)(\log {a\over b}+2)\geq 0$$ if we want real solutions. Can you finish it from this?
• Thank you for your hint! Continuing from you would get $$(\log\frac{a}{b})^2\ge4$$ where $\frac{a}{b}>0$ case 1:$$\log\frac{a}{b}\le-2$$ $$\frac{a}{b}\le10^{-2}$$ Hence $$0<\frac{a}{b}\le10^{-2}$$ case 2: $$\log\frac{a}{b}\ge2$$ $$\frac{a}{b}\ge100$$ – Trey Anupong Feb 4 at 17:14
Hint: It is $$(\ln(x))^2+\ln(x)(\ln(a)+\ln(b))+\ln(a)\ln(b)+1=0$$ it is a quadratic in $$\ln(x)$$.