The equation in question is $$\log_5x*(\log_5(2*\log_{10}a-x)*\log_x5+1)=2$$

Tried working this down with the rules of logarithms, got it down to a quadratic equation of $x$ with $a$ as one of its parameters, but I'm sure that's not the right way to do it because ultimately I would get a single value and not an interval.

Tried getting it down to a single logarithm but they get nested because of different bases. Trivial statements such as $a>0$ don't help (me at least).

I'm not really sure I have an idea of what I should do with this, any hints about what should I aim to get?

  • 1
    $\begingroup$ Did you check to see where the discriminant of the quadratic would be positive? $\endgroup$
    – JB King
    Apr 20, 2015 at 21:20
  • $\begingroup$ I'm pretty sure that I made a mistake along the way. What I've got doing that is that the $log_10(a)>0$. Why would that help me? $\endgroup$
    – John Doe
    Apr 20, 2015 at 21:26
  • 1
    $\begingroup$ If you have $\log_{10}a > 0$, you can conclude $a>1$. (But that's not a step I came to for this problem.) $\endgroup$
    – aschepler
    Apr 20, 2015 at 21:33
  • $\begingroup$ Why do you think you'll get a single value? Getting $\log_{10}(a)>0$ (I haven't checked if that's right) tells you a lot, because only some values of $a$ makes that true, and those values would be an answer. $\endgroup$ Apr 20, 2015 at 21:33
  • $\begingroup$ Well, wouldn't this basically tell me only $a>1$? I've got five possible answers and none of them are this simple, every answer includes $10^5$. $\endgroup$
    – John Doe
    Apr 20, 2015 at 21:36

1 Answer 1


Let's start by simplify your equation : \begin{align} \log_5 x\left [ \log_5\left(2\log_{10}a - x\right)\log_x 5 + 1 \right] = 2 &\Leftrightarrow \log_5\left(2\log_{10}a - x\right) + \log_5 x = 2\\ & \Leftrightarrow x \left(2\log_{10}a - x\right) = 25\\ & \Leftrightarrow -x^2 + 2\log_{10} (a) x -25 = 0 \end{align}

The discriminant of the last equation is : $$\Delta = 4 \log_{10}^2 a - 100.$$ The equation admits real solutions iff $\Delta \geq 0$ hence : $$4\log_{10}^2 a -100 \geq 0 \Leftrightarrow a\geq 10^5.$$ Edit : As @aschepler said the case $\log_{10} a \leq -5$ is rejected because one must have $2\log_{10}a -x >0$.

  • $\begingroup$ $\log_{10}a \geq 5$ satisfies the discriminant, but so does $\log_{10}a \leq -5$. (But the latter gets eliminated by the demand that $2 \log_{10} a - x > 0$.) $\endgroup$
    – aschepler
    Apr 20, 2015 at 21:47
  • $\begingroup$ @aschepler yeah I thought about it but I forgot to write it. Thanks! $\endgroup$
    – Joelafrite
    Apr 20, 2015 at 21:56

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