0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – dxiv Dec 14 '16 at 6:16
0
$\begingroup$

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 $x=65$

$\endgroup$
1
$\begingroup$

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

Hence $x=65$

$\endgroup$
1
$\begingroup$

\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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.