Simplify $\log_23\ \log_34\ \log_4 5\ \log_5 6\ \log_6 7\ \log_7 8$ How do I evaluate the product:
$$\log_23\ \log_34\ \log_4 5\ \log_5 6\ \log_6 7\ \log_7 8$$
I know that $$\log_ba=\frac{\log\ a}{\log\ b}$$
How can I apply it?
Thanks!
 A: You have everything that you need:
$$\log_n(n+1)=\frac{\log_2(n+1)}{\log_2n}\;,$$
so
$$\begin{align*}
\log_23\log_34\log_45\log_56\log_67\log_78&=\log_23\cdot\prod_{n=3}^7\left(\frac{\log_2(n+1)}{\log_2n}\right)\\
&=\frac{\prod_{n=3}^8\log_2n}{\prod_{n=3}^7\log_2n}\\
&=\log_28\\
&=3\;.
\end{align*}$$
A: Just apply the rule you stated and write the terms down , you will notice that all the terms cancel out except the denominator of the first and the nominator of the last one.
A: $$log_23\ log_34\ log_4 5\ log_5 6\ log_6 7\ log_7 8=\frac{\log 3}{\log 2}\frac{\log 4}{\log 3}\frac{\log 5}{\log 4}\frac{\log 6}{\log 5}\frac{\log 7}{\log 6}\frac{\log 8}{\log 7}=\frac{\log 8}{\log 2}=log_28=3$$
A: Hint:  $log_23=\frac{log3}{log2}, log_34=\frac{log4}{log3}$ and so on, find anything fishy?
A: Hint: Its $\frac{log 3}{log 2}\cdots \frac{log 8}{log 7}$
A: $log_23\ log_34\ log_4 5\ log_5 6\ log_6 7\ log_7 8$ 
= $\frac{log3}{log2} \times \frac{log4}{log3} \times \frac{log5}{log4} \times \frac{log6}{log5} \times \frac{log7}{log6} \times \frac{log8}{log7}$
log 4 = 2log 2 
log 8 = 3log 2 
By putting these values and simplifying you get : 3 as answer.
