Simplify $\log_23\ \log_34\ \log_4 5\ \log_5 6\ \log_6 7\ \log_7 8$

Question

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!

Answer 1

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.

Answer 2

Hint: Its $\frac{log 3}{log 2}\cdots \frac{log 8}{log 7}$

Answer 3

You have everything that you need:

$$\log_n(n+1)=\frac{\log_2(n+1)}{\log_2n}\;,$$

so

$$\begin{align*} \log_23\log_34\log_45\log_56\log_67\log_78&=\log_23\cdot\prod_{n=3}^7\left(\frac{\log_2(n+1)}{\log_2n}\right)\\ &=\frac{\prod_{n=3}^8\log_2n}{\prod_{n=3}^7\log_2n}\\ &=\log_28\\ &=3\;. \end{align*}$$

Answer 4

$$log_23\ log_34\ log_4 5\ log_5 6\ log_6 7\ log_7 8=\frac{\log 3}{\log 2}\frac{\log 4}{\log 3}\frac{\log 5}{\log 4}\frac{\log 6}{\log 5}\frac{\log 7}{\log 6}\frac{\log 8}{\log 7}=\frac{\log 8}{\log 2}=log_28=3$$

Answer 5

$log_23\ log_34\ log_4 5\ log_5 6\ log_6 7\ log_7 8$ = $\frac{log3}{log2} \times \frac{log4}{log3} \times \frac{log5}{log4} \times \frac{log6}{log5} \times \frac{log7}{log6} \times \frac{log8}{log7}$

log 4 = 2log 2 log 8 = 3log 2 By putting these values and simplifying you get : 3 as answer.

Answer 6

Hint: $log_23=\frac{log3}{log2}, log_34=\frac{log4}{log3}$ and so on, find anything fishy?