#Summation of logarithmic function
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I was trying to find a way to work with it with the log base change rule
but I cant really find a way
@agile mantle
it works with the log base change rule
change it all to base 3
and it all cancels out beautifully
hold on let me write it out
does it have to be base 3 or can I use any base
I think im losing u lmao
oh wait I think I see what u mean
so itl literally just be the last term over ln3
change it to the base of the first logarithm so there isnt an annoying denominator
should be ln (n-1) / lnn
yeah lol
oh
ig I dont think itl matter though no?
itl still be log base 3 of 6 at the end no?
yeah, log3(n)
why not just keep it as ln just to make it simple at the beggining
wouldnt it make now difference
because look at what the sum is
f(3^k)
log base 3 of that is quite easy to find
oh I see waht u mean
what*
yeah ig
I mean we get calculators anyways but yeah thats true
also would for example f(4) only be log3(4) or would it move up 4 in the sequence, ie; log3(4) * log5(6) * log6(7) * log7(8)
im assuming the first option
ik its kind of a dumb question
@slate crescent
yes
ok so the general formula is f(n) = ln(n)/ln(3)
yes
also just to confirm, the way I solved it from there was just brute force it
just add up all 20 of them
after knowing the shortcut for f(n)
or u could just look at it as 1 + 2 +3 + 4 +5 +6 +7 +8 +9 +10.... all the way to 20 ig
it’s 3 + … + 20
mby eah
there is a shortcut
just realized when i started writing it out lol
u can just arithmetic sequenece it
.
common difference is 1
ima be honest idk how u got that 23*9
buti m just gonna stick with n = 18, u1 = 3
d = 1
then just sum it
nvm I jsut got the same thing
whyd u write this though lol
helps see that it’s adding 23 up
each takes 2 terms, there are 18 terms, so 9 23s
+close