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I was studying logarithms and got stuck with this problem. Is there anyone who can solve this? Thank you in advance. The question - Given that $p$ and $q$ are positive and that $\ 4(\log_{10} p)^2+2(\log_{10} q)^2=9$ , find the greatest possible value of p such that the equality holds.

Edit

As per asker's note, the answer is:$10\sqrt 10$

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    $\begingroup$ For the greatest $p$, $4(\log p)^2$ (you might want to fix that typo) would also be the greatest. In order for that, we need to minimize the value of $2(\log q)^2$. Do you see something? 🙂 $\endgroup$
    – TheMather - or rather AMather
    May 27 at 7:24
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    $\begingroup$ The answer is $\ 10\sqrt10$ but I do not know how to solve that. $\endgroup$
    – Kyo
    May 27 at 7:44
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    $\begingroup$ Have you tried what I said in my comment? $\endgroup$
    – TheMather - or rather AMather
    May 27 at 7:47
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    $\begingroup$ Not yet. How can I minimize the value of $\ 2(log_{10} q)^2$. $\endgroup$
    – Kyo
    May 27 at 8:03
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    $\begingroup$ No problem! Hope it helps. $\endgroup$
    – TheMather - or rather AMather
    May 27 at 9:52

1 Answer 1

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The answer is really simple. First you have the equality $4(\log_{10}p)^2+2(\log_{10}q)^2=9$ Then, $9-4(\log_{10}p)^2=2(\log_{10}q)^2$ Now we factor the left hand side, $(3-2(\log_{10}p))(3+2(\log_{10}p))=2(\log_{10}q)^2$. Because the right hand side should always be positive, we deduce that the greatest value of $p$ is when: $2(\log_{10}p)=3$

This is because if $2(\log_{10}p)>3$, then the left hand side is negative. Therefore, $2(\log_{10}p)\leq3$ so 3 is the greatest value that the expression can take:

$\log_{10}p=\frac{1}{2}$

$p=10^{1.5}$

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    $\begingroup$ Thank you. That answered my question but I do not quite understand how you did that. $\endgroup$
    – Kyo
    May 27 at 8:06
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    $\begingroup$ Hmm... the right hand side can be zero, right? $\endgroup$
    – TheMather - or rather AMather
    May 27 at 8:23
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    $\begingroup$ Also, you're looking for the greatest $p$ so that $(3-2\log p)(3+2\log p)\geq0$, right? You might want to make that bit clearer 🙂 $\endgroup$
    – TheMather - or rather AMather
    May 27 at 8:28
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    $\begingroup$ @Haidara haha, I'm also a mobile user! Maybe you can just add some explanation. $\endgroup$
    – TheMather - or rather AMather
    May 27 at 8:32
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    $\begingroup$ Alright, nice! 🙂 $\endgroup$
    – TheMather - or rather AMather
    May 27 at 8:38

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