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

cartmanbrahhhhh
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2 Answers2

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I guess your question is related to solving the equation of $x$, and then you might be interested about some features/conditions about $x$. As I see it, if you assume for only real numbers, $x$ would always be positive if the solution exists. To see this:

$$\log (ax)\cdot\log(bx) + 1 = 0$$ $$\log(ax)\cdot\bigg(\log(ax\cdot\frac{b}{a})\bigg) + 1 =0$$ $$\log(ax)\cdot\bigg(\log(ax) + \log(b) - \log(a)\bigg) + 1 =0$$ $$\log^2(ax) + \bigg(\log(b) - \log(a)\bigg)\cdot \log(ax) + 1 =0$$

If we let $y=\log(ax)$, then we get $$y^2 +(\log(b) - \log(a))y + 1=0$$

which is a quadratic equation, whose solution is $\log(ax)=y=\frac{-(\log(a)-\log(b))\pm \sqrt{\Delta}}{2\times1}$, where $\Delta=(\log(b)-\log(a))^2-4\times1\times1.$

And you should be able to solve $\log(ax)$ if your $\Delta \ge 0$, and then you could get $x$. But as long as you get the solution, $x$ is garanteed to be positive, because you'll eventually have $\log(ax) = m$, and then $x = \frac{1}{a}e^m > 0$

But maybe you are interested in the condition that we have a solution for $x$, if that is case, we just need to make sure that $\Delta = (\log(b) - \log(a))^2 -4\ge 0$

That is $\log\frac{b}{a} \ge 2$ or $\log\frac{b}{a} \le -2$ for the solutions to exist; which means $\frac{b}{a} \ge e^2$ or $\frac{b}{a} \le \frac{1}{e^2}$.

Jay Zha
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  • This answer is very clear and helpful. Thank you.

    By the way, does this mean the question itself is incorrect? The question mentioned x > 0 but not f(x) > 0.

     – cartmanbrahhhhh May 18 '17 at 01:56
  • @cartmanbrahhhhh No problem! Could you accept it if you find it helpful :) – Jay Zha May 18 '17 at 01:57
  • I don't quite get it actually, first look I was able to understand the answer, how do i get the formula Δ = (log(b) - log(a))^2 - 4 \ge 0 ? – cartmanbrahhhhh May 18 '17 at 02:02
  • Thank you! Notice that we never talk $f(x)$ here, it's all about the solution $x$, I would not say the question is wrong, because $x>0$ could be understood as at least we want $x$, the solution to exist - if that is the case, then it's a valid question. What I stressed in my answer is that actually for this particular question, $x>0$ and $x$ exist are the same thing, because as long as $x$ exists, it is guaranteed that $x>0$. – Jay Zha May 18 '17 at 02:02
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    Ah I see. That clear some of the things up. – cartmanbrahhhhh May 18 '17 at 02:04
  • @cartmanbrahhhhh for the $\Delta \ge 0$ part: notice the last function is actually in the form of $ax^2+bx +c=0$, where you treat the $\log$ as a whole. Then are you familiar with the roots of this quadratic equation? i.e. $x=\frac{-b +/- \sqrt{\Delta}}{2a}$, where $\Delta = b^2-4ac$ – Jay Zha May 18 '17 at 02:06
  • Let us continue this discussion in chat. – cartmanbrahhhhh May 18 '17 at 02:07
  • I've updated my answer accordingly, and let me know if you have any more questions! – Jay Zha May 18 '17 at 02:14
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$\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 $a/b \le e^{-2}$.

marty cohen
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