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

Question

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!

Answer 1

HINT:

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?

Answer 2

Hint: It is $$(\ln(x))^2+\ln(x)(\ln(a)+\ln(b))+\ln(a)\ln(b)+1=0$$ it is a quadratic in $$\ln(x)$$.