anonymous
  • anonymous
\[\Huge \color{red}{LOG_ IN}\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
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ParthKohli
  • ParthKohli
Hmm -- that's actually very hard to read.
anonymous
  • anonymous
\[\large \log_12\log_21\log_31\log1_3...\log_{n-1}n=x\log _12\log_23\log_34...log _n(n-1)\]
anonymous
  • anonymous
is it ok now

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terenzreignz
  • terenzreignz
\[ \log_12\log_21\log_31\log1_3...\log_{n-1}n=x\log _12\log_23\log_34...log _n(n-1)\]
amistre64
  • amistre64
:) color is soo much better
terenzreignz
  • terenzreignz
What does it even matter, this part \[ \log_12\log_21\color{red}{\log_31}\log1_3...\log_{n-1}n=x\log _12\log_23\log_34...log _n(n-1)\] is already zero -.-
anonymous
  • anonymous
yes thats correct ...mistake
anonymous
  • anonymous
wait its actually last one \[\log_12\log_21\log_31\log_13...\log_1n=x\log _12\log_23\log_34...\log _n(n-1)\]
terenzreignz
  • terenzreignz
You still have a \(\large \log_31\) there...
terenzreignz
  • terenzreignz
Wait a minute... what is this even supposed to mean? \[\log_12\log_21\log_31\color{red}{\log_13}...\log_1n=x\log _12\log_23\log_34...\log _n(n-1)\] You know that that doesn't exist... -.-
anonymous
  • anonymous
sry this is what i want let me explain how i wanted to phrase this ?
terenzreignz
  • terenzreignz
Hit me
anonymous
  • anonymous
\[\log_12\log _21\log_13\log_31 ...\log_1(n)\log_n1\] such that if i use the property \[\log _ab=\frac{ 1 }{ \log _ba }\] hence \[\log_12\frac{ 1 }{ \log _12 }\log _13\frac{ 1 }{ \log _31 }...\log_1n \frac{ 1 }{ \log_1 n}=1\]
terenzreignz
  • terenzreignz
Why is there even a log base 1 ?
anonymous
  • anonymous
so i see\[\log_12=x \implies 1^x=2\implies \emptyset\]
terenzreignz
  • terenzreignz
yup
anonymous
  • anonymous
how can i phrase such a question to work is it \[\log_23\log_32\log_45\log_54...\log_n(n+1)\log_{n+1}n\]
terenzreignz
  • terenzreignz
These pairs (in respective colour) are all just equal to 1... \[\color{red}{\log_23\log_32}\color{green}{\log_45\log_54}...\color{blue}{\log_n(n+1)\log_{n+1}n}\]
anonymous
  • anonymous
yes exaclty
terenzreignz
  • terenzreignz
Well, that seems to be it for this particular mess of logs :P
anonymous
  • anonymous
DO YOU KNOW HOW TO DEAL WITH RIGHT HAND SIDE
terenzreignz
  • terenzreignz
right hand side also has a log base 1
anonymous
  • anonymous
IF YOU APPLY CHANGE OF BASE YOU HAVE \[\log _ab=\frac{ \log b }{ \log a }\] with the same reasoning we only have\[nx=1\]
anonymous
  • anonymous
of course i will have to start again at an interger more than one
terenzreignz
  • terenzreignz
right
terenzreignz
  • terenzreignz
uhh...
terenzreignz
  • terenzreignz
The RHS is daunting... might need to get a second opinion on that :)

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