<s>[INST] what is the value of $ x$? If $\log_{2}3^4\cdot\log_{3}4^5\cdot\log_{4}5^6\cdot....\log_{63} {64}^{65}=x!$, what is the value of $ x$?
I've tried
$$\log_2 3^4\cdot\log_3 4^5\cdot\log_4 5^6\cdot.... \log_{63} 64^{65}$$
$$=\dfrac{4\log 3}{\log 2}\cdot\dfrac{5\log 4}{\log 3}\cdot\dfrac{6\log 5}{\log 4}\cdots\dfrac{65\log 64}{\log 63}$$
don't know how to solve futher steps, please help.
Thanks
 [/INST] Write    $\;\;\;\;\;\;\;\dfrac{4\log 3}{\log 2}\cdot\dfrac{5\log 4}{\log 3}\cdot\dfrac{6\log 5}{\log 4}\cdots\dfrac{65\log 64}{\log 63}\;\;\;\;\;$  as
$$=\dfrac{4\log 3}{\log 2}\cdot\dfrac{5\log 4}{\log 3}\cdot\dfrac{6\log 5}{\log 4}\cdots\dfrac{65\cdot 6\log 2}{\log 63}$$
$\{\because \log64=\log2^6=6\log2\}$      
Every logarithmic part cancels out, and we have
$$=4\cdot 5 \cdot 6 \cdots \color{red}{6}\cdot 65$$
$$=\color{red}{2\cdot 3} \cdot 4\cdot 5\cdot 6 \cdots 65$$
$$=65!$$
Hence $x=65$
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