Find the value of $\log_8 9 \times \log_9 10 \times \cdots \times \log_n(n+1) \times \log_{n+1}8$ I'm completely lost on this question. I've been Googling around to no success. 

Find the value of $$\log_8 9 \cdot \log_9 10 \dotsm  \log_n(n+1) \cdot \log_{n+1}8$$

I'm completely stumped as to where to even begin looking. If someone could point me in the right direction I'd appreciate it.
 A: Remember that $\log_a b = \frac{\ln b}{\ln a}$. The product then telescopes.
A: Put each logarithm in the same base:
$$\log_89\cdot\log_910\cdot\cdots\cdot\log_n(n+1)\log_{n+1}8=\frac{\log 9}{\log 8}\cdot\frac{\log 10}{\log 9}\cdot\cdots\cdot\frac{\log(n+1)}{\log n}\frac{\log 8}{\log (n+1)}=1.$$
A: Rewrite them as
\begin{align}
\frac{\color{red}{\log\left(9\right)}}{\log\left(8\right)}\frac{\color{green}{\log\left(10\right)}}{\color{red}{\log\left(9\right)}}\frac{\color{blue}{\log\left(11\right)}}{\color{green}{\log\left(10\right)}}\frac{\log\left(12\right)}{\color{blue}{\log\left(11\right)}}\cdots\frac{\log\left(n+1\right)}{\log\left(n\right)}\frac{\log\left(8\right)}{\log\left(n+1\right)}=1,
\end{align}
since all of the logs cancel in-between and that leaves the two $\log\left(8\right)$'s on either side in the denominator and numerator.
A: Hint:
$$\log_b x = \frac{\log_k x}{\log_k b}.$$
Use this and you'll see that the answer is $1$.
A: There is a litle-known and under-appreciated "cancellation law" for logarithms:
$$log_a(b) \cdot log_b(c) = log_a(c)$$
(This is essentially the same as the "change-of-base" formula, written multiplicatively instead of using division.)
If you apply that property here you'll see that everything in the problem cancels out.
A: Hint: $$\log_b a\times \log_c b=\log_c a$$
A: Take the leftmost item and multiply it by the rightmost item:


*

*$\log_{8}{9}\times\log_{n+1}{8}=\log_{n+1}{8}^{\log_{8}{9}}=\log_{n+1}{9}$


Take the leftmost item and multiply it by the previous result:


*

*$\log_{9}{10}\times\log_{n+1}{9}=\log_{n+1}{9}^{\log_{9}{10}}=\log_{n+1}{10}$


Take the leftmost item and multiply it by the previous result:


*

*$\dots$


Take the remaining item and multiply it by the previous result:


*

*$\log_{n}(n+1)\times\log_{n+1}{n}=\log_{n+1}{n}^{\log_{n}(n+1)}=\log_{n+1}(n+1)=1$

A: People always go with logs rather than powers. Often these questions are easier to do with powers rather than logs. Note that $$8^{\log_89}=9$$ It is then obvious how things develop if we raise $8$ to the power of the given expression (which I'll call $x$) to get $$8^x=(n+1)^{log_{n+1}8}=8$$ whence $x=1$
A: Hint:
$$ a= b^{\displaystyle\log_ba}$$
$$\log_c a = \log_c b^{\displaystyle\log_ba}$$
$$\dots$$
