# 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

• Hint: it "telescopes". Try writing a few more terms at the beginning and end, then start canceling out $\log$ terms between numerators and denominators. See what's left. – dxiv Dec 14 '16 at 6:16

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 $x=65$

You get $$4\cdot 5\cdot 6\cdots 65\cdot {\log 64\over \log 2}\\=1\cdot2\cdot3\cdot4\cdot5\cdots65=x!$$

Hence $x=65$

\begin{align}\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}&=\frac{65!}{3!}\frac{\log 64}{\log 2} \\&=\frac{65!}{3!}\frac{\log 2^6}{\log 2} \\ &=\frac{65!}{3!}\frac{\log 2}{\log 2}.6 \\&=65! \end{align}