Question

@Kainui

Answer 1

So like try to tell me what you know and understand about logs so I can fill in the gaps and no explain stuff you already know if that's possible

Answer 2

I know the basic properties

Answer 3

lol

Answer 4

hey start if I dont understand smth ill stop u and question

Answer 5

I guess I could say this is true and you would believe me cause you memorized it \[\Large \log_{10}10=1\] but we are really starting from \[\Large 10^1=10\] in a kind of way.

Like for example, you know that if you want to solve this equation \[\Large y=b^x\] You wuld just do \[\Large \log_b y = x\] So it only makes sense by substituting y=b^x into this we have: \[\Large \log_b b^x = x\] and not only that, log rules are really just the same as exponent rules if you think about it.

Answer 6

Ya this stuff im familiar with

Answer 7

Well that last line is what you wanted to know I thought haha.

Answer 8

hahaha nooo

Answer 9

why does
\[\large x^{\log_x(y)}=y\]

Answer 10

ohhh waitttt

Answer 11

hahaha i get it

Answer 12

Haha I plugged it in backwards so like instead of \[\Large f(f^{-1}(x))=x\] I showed \[\Large f^{-1}(f(x))=x\] lol

Answer 13

So depending on how you plug 'em in to each other \[\Large y=b^x \\ \Large \log_b y = x\] gets you either \[\Large \log_b b^x = x \\ \Large b^{\log_bx}=x\]

Answer 14

Wow OS is super laggy for me.

Answer 15

yaa yaaa now i see it

Answer 16

Thanks :D

Answer 17

ya u need a new comp btw

Answer 18

Cool yeah I am actually trying to work on something with logs right now but I can't figure it out

Answer 19

hahahah doubt i can be off assistance

Answer 20

Im still a tweenie weenie

Answer 21

Yeah it's sort of hard to describe it could take a while to even explain what I am trying to do, unless you're familiar with how fourier series coefficients are found. =P

Answer 22

k no im not ... ya need dan for that