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What is the value of $\left(\log_{21}(3)\right)^2+\log_{21}(147)\log_{21}(1323)$ ?

$1)1\qquad\qquad2)2\qquad\qquad3)3\qquad\qquad4)4$

To solve this question I tried writing the expression as:

$$\left(\log_{21}(3)\right)^2+\log_{21}(7^2\times3)\log_{21}(7^2\times3^3)$$ $$=\left(\log_{21}(3)\right)^2+(2\log_{21}7+\log_{21} 3)(2\log_{21}7+3\log_{21}3)$$

I don't know how to continue.

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    $\begingroup$ Don't be so eager to separate the $\log7$ from the $\log3$. The most potent course of simplification you have here is $\log21=1$. $\endgroup$
    – Arthur
    Jul 1 at 10:23
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    $\begingroup$ @Arthur Thanks! The question looks horrible for a timed exam and I became a little nervous! $\endgroup$ Jul 1 at 10:29
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    $\begingroup$ Denote $x=\log_{21}3$ and $y=\log_{21}7$. Then, $147=21\times 7$ and $1323=21^2\times 3$; the original expression is then $x^2+(1+y)(2+x)=x(x+y)+(x+y)+y+2$; iteratively use $x+y=1$ to simplify. $\endgroup$ Jul 1 at 10:47
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    $\begingroup$ @Prasun Biswas: the computation is simpler if one notices that $147=21^2/3$ and $1323=21^2 \times 3$. $\endgroup$
    – Mindlack
    Jul 1 at 10:51
  • $\begingroup$ @Mindlack: Nice! Indeed, that is far more simpler because we wouldn't need the $y$, we have $x^2+(2-x)(2+x)=x^2+4-x^2=4$ Somehow, I missed that. +1 $\endgroup$ Jul 1 at 10:55
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Here’s a quick-and-dirty solution based on the fact that one of the answers must be correct.

All the quantities are positive. Since $1323>441=21^2$, its base-$21$ logarithm is greater than $2$. As $147 \geq 21 \times 5$, its base-$21$ logarithm is at least $1.5$. This means that the sum is greater than $2 \times 1.5$ so it must be $4$.

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  • $\begingroup$ I don't get the second point regarding $147 \geq 21 \times 5$. Could you elaborate? $\endgroup$
    – user71207
    Jul 1 at 11:04
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    $\begingroup$ @user71207 $147\ge21\times5\ge21\sqrt{21}$ And the fact that $147\ge21^{\frac32}$ means $\log_{21}(147)\ge\frac32$ $\endgroup$ Jul 1 at 11:09
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$\log_{21}1323=\log_{21}(3^3\cdot 7^2)=\log_{21}3+\log_{21}21^2=2+\log_{21}3$
$\log_{21}147=\log_{21}21+\log_{21}7=1+\log_{21}7$

$$(\log_{21}3)^2+\log_{21}147\cdot\log_{21}1323 \\=(\log_{21}3)^2+(1+\log_{21}7)(2+\log_{21}3) \\=\log_{21}3(\log_{21}3+\log_{21}7)+\log_{21}3+2\log_{21}7+2 \\=\log_{21}3(\log_{21}3+\log_{21}7)+(\log_{21}3+\log_{21}7)+\log_{21}7+2 \\=\log_{21}3+1+\log_{21}7+2 \\=4$$

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  • $\begingroup$ Hi, the problem in this post and also this one that I asked were from entrance exam in the country I'm living (average time for solving each question is about 90 seconds). You solved both of them very well and I think you are very good at solving these questions! So can you please suggest some books or name of exam that you think are good for an exam that contains these type of questions? Thanks in advanced! $\endgroup$ Jul 27 at 19:58
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    $\begingroup$ I mainly use books that are meant for olympiads, so I don't know many books that are meant for solving questions fast. $\endgroup$
    – Asher2211
    Jul 28 at 6:28
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Alternate:

Let $L(x)$ denote $\log_{21}(x).$

Then $[L(3)]^2 + [L(147) \times L(9 \times 147)]$

$= [L(3)]^2 + ~\langle ~L(147) \times ~\{ [2 \times L(3)] + L(147) ~\} ~\rangle$

$= [L(3)]^2 + [L(147)]^2 + [2 \times L(3) \times L(147)]$

$= [L(3) + L(147)]^2.$

At this point, you have that $3 \times 147 = 21^2$,

so $[L(3) + L(147)] = 2.$

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