Further more, now we can represent the exponential and logarithmic forms as b^x = m ---> log_b m = x, where (b) represents the base number, (x) is a variable, and (m) is the result. Since any number (where b ≠ 0) (b)^0 = 1, there exists log_b 1 = 0. Also any number (b)^1 = b ---> log_b b = 1.
You will find through a time consuming and lengthy process, the exponential constant, I wish to not write it all down right now, but just know that 2^x = m --> log_2 m = x and 3^x = m ---> log_3 m = x and there exist a constant (e) that sits between 2 and 3 (≈ 2.71...), found by a growth calculation (1 + 1/n)^n. This will help later on with cancelations and such for calculations, (Like how you can cross cancel rational number in multiplication). Now there exists e^x and ln x. Where ln x is also just log_e e = x.
The author is putting this on hold for now as this page ends an arc, and he is switching and updating Sultry Summer once he finishes an arc for that, he will come back here
Add new comment
Guys help me learn logarithms . Will pay u absolutely nothing
aight so you logarithms are the opposite of exponentiating. So if you do 3^10 then log base 3 of that will be 10. It undoesz the 3^.
if you wanna learn the basics probably look up eddie woo or some math mf on yt
yes, in other terms:
log_a(b) = the number of times you have to moltiplicate a to get b:
(e.g):
log_2(8) = 3
because 2^3=2×2×2=8
so we had to moltiplicate 2 for 3 times to get 8!
hope this is easy to understand!
Further more, now we can represent the exponential and logarithmic forms as b^x = m ---> log_b m = x, where (b) represents the base number, (x) is a variable, and (m) is the result. Since any number (where b ≠ 0) (b)^0 = 1, there exists log_b 1 = 0. Also any number (b)^1 = b ---> log_b b = 1.
You will find through a time consuming and lengthy process, the exponential constant, I wish to not write it all down right now, but just know that 2^x = m --> log_2 m = x and 3^x = m ---> log_3 m = x and there exist a constant (e) that sits between 2 and 3 (≈ 2.71...), found by a growth calculation (1 + 1/n)^n. This will help later on with cancelations and such for calculations, (Like how you can cross cancel rational number in multiplication). Now there exists e^x and ln x. Where ln x is also just log_e e = x.
Correction : ln(x) <---> log_e(x)
So now, whenever you are asked to solve for x, just know that ln(e^x) = x and e^lnx = x.
I'm sure this can be explained better, and I humbly request further support in assisting this person in their journey through logarithms.
But hopefully you get the gist.
(8/10) Incog do be cogging
Can't wait to see the new pages tonight
Me dik cum
May Doc have mercy on your soul
You cannot be talking like that when you clicked on this website.
Well fall is almost here so this will soon be back up and running, and then we wait again for sultry summer.
this real early sex education and hands on experence too
Part 3
Lol may the Lord forgive us is a tag
Now to continue with the original
The author is putting this on hold for now as this page ends an arc, and he is switching and updating Sultry Summer once he finishes an arc for that, he will come back here
Don't expect this to be updated for a good while
fuck
So... p3?
Can't she move faster than sound? she could have easily made it outside in that timeframe
is this even legal
Probably. Idk
Are you legally allowed to be here?
Are you legally allowed to be here?
don't be such a stupid human being and go search the results for yourself
A random administrator picked this pfp for me. I didn't expect something so honorable...
IM KEEPING THIS BITCH. YOU CANT DO ANYTHING TO STOP ME
To be continued, now back to Sultry Summer
You too huh
Pages