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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!

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

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    $\begingroup$ 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: $$\log\frac{a}{b}\ge2$$ $$\frac{a}{b}\ge100$$ $\endgroup$ – Trey Anupong Feb 4 at 17:14
  • $\begingroup$ Nice, I am glad I could help and you could finish it. $\endgroup$ – user376343 Feb 4 at 18:12
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Hint: It is $$(\ln(x))^2+\ln(x)(\ln(a)+\ln(b))+\ln(a)\ln(b)+1=0$$ it is a quadratic in $$\ln(x)$$.

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  • $\begingroup$ It is very useful, thank you for your hint! $\endgroup$ – Trey Anupong Feb 4 at 19:01

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