Question

2041^2n=243^n+2

Answer 1

what is the question :)

Answer 2

uhhh, find n?? :P

Answer 3

yes

Answer 4

i'm gonna use log to solve this, is that ok?? it's the weirdest method, i know :P

Answer 5

Log would be helpfil

Answer 6

i have NO clue of what i've done, (lol) btu the answer i've got is 1.127. Should i show the working??

Answer 7

yes

Answer 8

so after putting log on both sides, and then you bring down the indices on both sides, and you get

\[2n \log_{}2041 = n+2\log_{}243\]

Answer 9

Take 2log2041 common

Answer 10

take out the value of log 2041

Answer 11

This question is only possible by using logs

Answer 12

\[
2 n \log (2014)=(n+2) \log (243) \\
2 n \log (2014)-n \log (243)=2
\log (243)\\
n (2 \log (2014)-\log (243))=2
\log (243)\\
n=\frac{2 \log (243)}{2 \log
(2014)-\log (243)}=1.12995
\]

Answer 13

then you do

\[\frac{\log_{}2041}{\log_{}243} = \frac{n+2}{2n}\]

Answer 14

then 2.775n = n+2, and then you get n as what i got above :P

Answer 15

@tanvidais13 did you see how I did it?

Answer 16

ok thnks

Answer 17

yupss, i did, it's just i'd typed the whole thing out before, and then forgot to post it :P and then posted in 10 years later. and your method seems fin too :D

Answer 18

*fine.