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Honestly I've never been good with logarithms, they mess my head.

In the equation; $$\log_{10}(ax)\log_{10}(bx)+1=0$$

with $a>0$, $b>0$ and $x>0$

This equation only has solutions when the ratio $b/a$ is in the interval $(0,m]$ or the interval $[M,\infty)$. I need to find these intervals (i.e the $m$ and the $M$).

I don't know how to answer this, I tried to expand the expression and change 1 to the right adding the base then to $-1$ and try to solve from there but it was useless the far I've been is ;

$$\log_{10}(a)\log_{10}(b)+\log_{10}(a)\log_{10}(x)+\log_{10}(b)\log_{10}(x)+\log_{10}(x)^2+1=0$$

Selrach Dunbar
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Marco
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    You should type math with dollar signs, like this:$a<b$ for $a<b$, and use the underscore for subscripts, like $\log_{10}(ax)$ for $\log_{10}(ax)$. – Joe Jun 06 '21 at 02:34
  • Thanks. I'll try to make the changes. – Marco Jun 06 '21 at 02:41
  • What is the task here? Does the cryptic notation >=a/b> mean that you want to find the ratio $\frac ab$ as a function of $x$? That sounds somewhat unlikely -- if you have $a,b,x$ that satisfy your equation, then interchanging $a$ and $b$ will also satisfy your equation but it will give a different $\frac ab$. – Troposphere Jun 06 '21 at 03:30
  • What does $>=a/b>$ mean? – Arctic Char Jun 06 '21 at 03:44
  • Trying to find the intervals – Marco Jun 06 '21 at 03:47

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Hint

Using $$A=\log_{10}(a)\qquad B=\log_{10}(b) \qquad X=\log_{10}(x)$$ you last equation write $$ X^2+(A +B) X+(A B +1)=0$$ If it has a root, then ....

Claude Leibovici
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