# The product of logarithms

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:$$10\sqrt 10$$

• 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? 🙂 May 27 at 7:24
• The answer is $\ 10\sqrt10$ but I do not know how to solve that.
– Kyo
May 27 at 7:44
• Have you tried what I said in my comment? May 27 at 7:47
• Not yet. How can I minimize the value of $\ 2(log_{10} q)^2$.
– Kyo
May 27 at 8:03
• No problem! Hope it helps. May 27 at 9:52

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:

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

$$p=10^{1.5}$$

• Thank you. That answered my question but I do not quite understand how you did that.
– Kyo
May 27 at 8:06
• Hmm... the right hand side can be zero, right? May 27 at 8:23
• 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 🙂 May 27 at 8:28
• @Haidara haha, I'm also a mobile user! Maybe you can just add some explanation. May 27 at 8:32
• Alright, nice! 🙂 May 27 at 8:38