Question

log in

Answer 1

What's the question matey?

Answer 2

\[ {\log_12\log_2 3\log_34...\log_n(n+1)}=\frac {x}{ \log _{n+1}n^n}\]

Answer 3

find x in terms of n

Answer 4

big version\[\huge \color{blue}{\log_12\log_23\log_34...\log_n(n+1)}=\frac{x}{\log _{n+1}n^n}\]

Answer 5

i just made this up,already know how to do it
can you see the trick

Answer 6

is that multiplying a bunch of logs together?

Answer 7

yes

Answer 8

im curious if just multiply bot sides by that log denominator thing would suffice; that gets all ns to one side and all xs to the other.

Answer 9

ye that will do but the LHS can be simplified without multiplying

Answer 10

i dont have the wherewithal at the moment to try to sort thru these cobwebs to simplify that :)

Answer 11

wait can i edit the question right hand side

Answer 12

\[\frac{x}{n \log _{n+1}1}\]

Answer 13

HINT change of base

Answer 14

@amistre64

Answer 15

using \[\log _ab=\frac{\log b}{\log a}\]
\[\frac{ \log 2 }{ \log 1 }\frac{ \log 3 }{ \log 2 }\frac{ \log 4 }{ \log 3 }...\frac{ \log n }{ \log n-1 }\frac{ \log n+1 }{ \log n }=\frac{x}{n \log _{n+1}n}\]
hence we have all logs from 2 to n cancel out
\[\frac{\log n+1}{\log 1}=\frac{x}{n \log _{n+1}1}\]
\[n \frac{\log n+1}{\log 1}\frac{1}{\log n+1}=x \implies x=n\]