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Trying to evaluate $(\log_2(3)+\log_4(9))(\log_3(2)+\log_9(4))$ without a calculator
Using the change of basis formula, It can be seen that $\log_4(9)=\log_2(3)$ and $\log_3(2)=\log_9(4)$
We can now express our original question in a few different ways, for example:
$(\log_2(3)+\log_4(9))(\log_3(2)+\log_9(4))$
$(\log_2(3)+\log_2(3))(\log_3(2)+\log_(2))$
However, I still have not been able to evaluate this… Help appreciated!!
Best Answer
If $x=\log_9(4)$ then $9^x=4$ so $x<1$, but if $x=\log_4(9)$ then $4^x=9$ so $x>1$. So your assertion $\log_4(9)=\log_9(4)$ cannot be correct.
We do, however, want to use a change of basis formula with $\log_4(9)=\log_2(3)$ and $\log_9(4)=\log_3(2)$:
$$ \begin{align}(\log_2(3)+\log_4(9))(\log_3(2)+\log_9(4))&=(2\log_2(3))(2\log_3(2))\\ &=4\log_2(3)\log_3(2)\end{align}$$
If $x=\log_2(3)$ and $y=\log_3(2)$ then $2^x=3$ and $3^y=2$, so $2^{xy}=(2^x)^y=3^y=2$ and thus $xy=1$.
(You might have seen this as a general "log rule", $\log_a(b)=\frac{1}{\log_b(a)}$.)
So $4\log_2(3)\log_3(2)=4xy=4$.