precalculus

[steel venture]

well you know what ! means right?

like 6!

[novel dirge]

yes

that would be 6x5x4x3x2x1

[steel venture]

ok

so if you have x!/(x-1)! = x

does that make sense

[novel dirge]

is that an example or should I do that in my problem?

[steel venture]

just walking you through it

[novel dirge]

oh, ok

I understand

[steel venture]

then if we have x!/(x-2)!

what do we get

[novel dirge]

ok, I don't understand that

wanted to say I understand that it is an example...

how do we get that it is equal to x?

[steel venture]

x! is x* x-1 *x-2 *x-3... 3 * 2 * 1

right

x-1! is the same but without the x

so when we divide all we have left is x

did you follow that

[novel dirge]

Yeah

kinda

I understand how we got that now

but not sure how to apply it to other examples

[steel venture]

lets keep moving

now instead of x-1!

we do (x-2)! at the bottom

so instead of x-1 *x-2 *x-3... 3 * 2 * 1

its x-2 *x-3... 3 * 2 * 1

so when we divide x! by (x-2)!

we get x * (x-1)

do you see that

[novel dirge]

so (x-1)!

would be (x-1)-1 * (x-1)-2 ...

?

[steel venture]

no

the number inside the parentheses is the number you start at

so (x-1)! is (x-1) * (x-2) * (x-3) ... 3 * 2 * 1

(x-5)! is (x-5) * (x-6) * (x-7) ... 3 * 2 * 1

(x+115)! is (x-115) * (x-116) * (x-117) ... 3 * 2 * 1

(2x)! is (2x) * (2x-1) * (2x-2) ... 3 * 2 * 1

do you get it

[novel dirge]

yes

[steel venture]

ok

so in the left side of this example

https://media.discordapp.net/attachments/363224154469826562/694083225370230794/JPEG_20200330_091820.jpg?width=346&height=702

x!/(x-2)! = [x * (x-1) * (x-2) ... 3 * 2 * 1] / [(x-2) * (x-3) * (x-4) ... 3 * 2 * 1]

right?

whoops its x-3

x!/(x-3)! = [x * (x-1) * (x-2) ... 3 * 2 * 1] / [(x-3) * (x-4) * (x-5) ... 3 * 2 * 1]

[novel dirge]

I am rewriting this in my notebook so I can understand it easier

[steel venture]

good

[novel dirge]

hard to read for me in this format

[steel venture]

yeah do what you need to

sorry i cant show it any better

[novel dirge]

np

thanks for trying to explain

[steel venture]

eah ofc

do you get what i sent though? after you write it down

[novel dirge]

still writing it down

[steel venture]

yeah lmk

[novel dirge]

had to find a pen

did you intentionally let out the 3! ?

[steel venture]

which 3!

[novel dirge]

on the left side

[steel venture]

oh yes

its just a constant

we can worry about it later

[novel dirge]

it is 3!*(x-3)!

oh

ok

wrote it down

I understand how we got that

but can I write it like that?

the *... * 3* 2* 1

[steel venture]

gimme 1 sec

brb

[novel dirge]

ok

[steel venture]

ok

yeah

just dont use asteriscs

use dots or the multiply sign

lol

but its just for visual understanding

after you get that

x!/(x-3)! = [x * (x-1) * (x-2) ... 3 * 2 * 1] / [(x-3) * (x-4) * (x-5) ... 3 * 2 * 1]

this

can be simplified to

x * (x-1) * (x-2)

[novel dirge]

how exactly?

I understand that it can be simplified

but not sure how

[slender bay]

Maximo incoming..just ping him/her

[novel dirge]

@steel venture like this?

[steel venture]

like what

[novel dirge]

that question was fo the cup

[steel venture]

oh lol

[slender bay]

Yeah..

[steel venture]

oh ok let me explain

[slender bay]

Like that

[novel dirge]

and I asked you if you can explain it a bit more how to simplify that

[steel venture]

x!/(x-3)! = [x * (x-1) * (x-2) ... 3 * 2 * 1] / [(x-3) * (x-4) * (x-5) ... 3 * 2 * 1]

we know this

[novel dirge]

yes

[steel venture]

now you can see that [x * (x-1) * (x-2) ... 3 * 2 * 1]

this part here

can be further expanded into'

[x * (x-1) * (x-2) * (x-3) * (x-4) * (x-5) ... 3 * 2 * 1]

do you see it?

[novel dirge]

yes

[steel venture]

[(x-3) * (x-4) * (x-5) ... 3 * 2 * 1] and that right side looks exactly like this

[(x-3) * (x-4) * (x-5) ... 3 * 2 * 1]

x!/(x-3)! = [x * (x-1) * (x-2) ... 3 * 2 * 1] / [(x-3) * (x-4) * (x-5) ... 3 * 2 * 1]

do you see where i got it from

[novel dirge]

yes

then the top and bottom cancel out

and all that is left is from x to x-4

[steel venture]

yep

no

x to x-2

[novel dirge]

yeah

my bad

went the wrong way

[steel venture]

npnp

[novel dirge]

still not used that this goes the other way around

[steel venture]

yeah it can be a lil confusing now and then

but anyhow

you're getting it right

[novel dirge]

yes

I now do understand how to do this part

[steel venture]

ok so now we have
[x * (x-1) * (x-2)]/3!

[novel dirge]

but I still have to add the 3! to it

[steel venture]

on the left side

yep

[novel dirge]

what now?

and what to do on the right side where I would have 4!((x+2)-4)!

do I just write that as 4!(x-2)! and do it like the first one?

[steel venture]

forget about the four for now

do it just like the first one and lmk what you get

[steel venture]

awesome

you got it perfectly

now its a matter of simplifying

if you don't know where to go for the next step lmk

[novel dirge]

yeah

I don't know

[steel venture]

do you see any common terms on either side

[novel dirge]

I know I can write out the 3 and 4

X-1

And x

[steel venture]

so

what can you do

[novel dirge]

Maybe multiply the whole equation with those 2?

[steel venture]

or you could

divide

[novel dirge]

yes

that

I want to get it to be on the bottom

and can I switch the sides ?

[steel venture]

well ok before you do anything

[novel dirge]

put the right part to the left

[steel venture]

its just two big multiplications right?

[novel dirge]

what exactly?

[steel venture]

the left and right sides

[novel dirge]

this or what I wanted to do?

I guess

[steel venture]

ok so if both sides have an x term being multiplied to them

what can you instantly do

to simplify

[novel dirge]

divide by that term

wo if we have x/2 we want to get x/(2*x)

[steel venture]

yeah

which is equal to what

[novel dirge]

1/2

[steel venture]

so

with the two common terms

try that out and see where you end up

[novel dirge]

now I just do the same with x and x-1 ?

[steel venture]

yep

also since its an equation

you can switch sides as you please

[novel dirge]

,rccw

[steel venture]

um ok from this point forwards you could expand the right and divide by (x-2)

but then you'd have to divide terms and idk if you know how

[analog relic]

@harsh cipher uh, $3=3(1)$ since $1=\log_a(a)$ where a can be anything, $3(1)=3\log_2(2)=\log_2[(2)^{3}]=\log_2(8)$

[obsidian monolith]

ιχιση_тαкαgι:

[novel dirge]

@steel venture what do you mean by expand it?

[steel venture]

foil

[novel dirge]

I don't have anything to expand anymore

can only multiply it

[steel venture]

(x+2)(x+1)

[novel dirge]

but I still can't get an (x-2) out of it

yea

[steel venture]

yeah you can divide both sides by (x-2)

and do long division

[novel dirge]

that is what I want to try, but I don't see how I can get a negative from 2 positives

didn't learn long division

[steel venture]

ah

give me 2 seecs

[novel dirge]

ok

[steel venture]

so instead of that

just expand the right side

and multiply by 3!

[novel dirge]

so I can also just shift it to the left

and do basic addition

where I find the comon bottom part

which would be 3! * 4!

right?

[steel venture]

yeah

[novel dirge]

ok

@steel venture

[steel venture]

looks good

although you gave the value for x

you dont need x e r do ou

you

[analog relic]

$^{x}C_{3}=^{x+2}C_{4}$
$\implies \frac{x!}{(3!)(x-3)!}=\frac{(x+2)!}{(4!)(x-2)!}$
$\implies \frac{(x)(x-1)(x-2)}{3!}=\frac{(x+2)(x+1)(x)(x-1)}{4!}$
$\implies \frac{x-2}{3!}=\frac{(x+2)(x+1)}{4!}$
$\implies x-2=\frac{(x+2)(x+1)}{4}$

[obsidian monolith]

ιχιση_тαкαgι:

[analog relic]

Then idk alr

[novel dirge]

you dont need x e r do ou
@steel venture ?

[steel venture]

X e R

it just says that x is an element of real numbers

however you already gave the value

unless thats a negative in the square root

[analog relic]

Ya

[novel dirge]

@analog relic I don't understand the last part

how did you get rid of the 3! and ! in the other part?

[analog relic]

3!=32=6
4!=43*2=24

[novel dirge]

yes

[analog relic]

Ya

[novel dirge]

I did that, too

[analog relic]

So denominator cancel out ah

[novel dirge]

how?

first time seeing that

and is what I did good? can I turn that in?

[analog relic]

$\frac{x}{6}=\frac{x}{24}$ for instance

[obsidian monolith]

ιχιση_тαкαgι:

[analog relic]

So 6 is the highest common multiple of 6 and 24

So can cancel 6 from both denominators

$\frac{x}{1}=\frac{x}{4}$

[obsidian monolith]

ιχιση_тαкαgι:

[analog relic]

Just common denominator

[steel venture]

yo how do you do the texit stuff so quick? is it just hardwired into your brain

[analog relic]

Idk, got used to it after awhile

I love this stuff

Cause look very presentable

[steel venture]

absolutely

[analog relic]

There's a website

[steel venture]

i just dont have the drive to do it

[analog relic]

To show you the codes

[novel dirge]

I wanted to learn laTex but don't have the brain power for it

[analog relic]

http://detexify.kirelabs.org/classify.html

You draw the symbol and it will show you the code

So I took some time to learn it

It's very nice looking on typing

[novel dirge]

@steel venture @analog relic so, can I send what I did to the teacher?

[steel venture]

i was learning latex for a bit but i didnt in the end

i think its good yeah

[novel dirge]

great

thank you

[analog relic]

Uhh ya sure

But x has t be a real whole positive number

to*

That's where Im stucl

stuck

[steel venture]

yeah same

since its combinatorics

[analog relic]

Cause negative square root is not a real number

[steel venture]

nor a whole one

[analog relic]

Ya

That's why

[novel dirge]

that is what I am confused about, too

[steel venture]

maybe say that the solution d.n.e?

[warped dagger]

how would i solve for y if cos(37) equals 0,7986, any point in the right direction is appreciated

[novel dirge]

is the process correct?

[steel venture]

the process is perfect pancake

i think at least

[novel dirge]

that is important

[steel venture]

its just the answer

[analog relic]

Ya I think the process is ok

It's the answer

Ya

[novel dirge]

I also just tried solving it with photo math

[steel venture]

@warped dagger isolate the y

[novel dirge]

gives the same answer

x € R

[steel venture]

aye just say x D.N.E.

[novel dirge]

not €

[analog relic]

Also, $(x+2)(x+1)=x^{2}+3x+2$

[warped dagger]

@steel venture so i just solve it like a normal equation?

[obsidian monolith]

ιχιση_тαкαgι:

[novel dirge]

x /€ R

[steel venture]

@warped dagger i don't see why not

[warped dagger]

Thanks, that popped up to me aswell, but wasnt sure

[novel dirge]

@analog relic damn

idk how I did that

I even said that I can't get negatives from all positives

[analog relic]

It's ok

Just a careless mistake, once you do over and over, you will suddenly stop making such mistakes ah

Or the frequency of the mistakes will be lesser and lesser

[novel dirge]

yeah

but even on exams

I almost always fail on the easiest problems

because of this

[analog relic]

Exam is high stress so normally, you would focus on the big picture and not little details

Causing this

[novel dirge]

wrong multiplication or something like that

[analog relic]

Ya

[novel dirge]

solve the hardest problems perfectly and mess up on the easy ones

[analog relic]

Of course (same)

[steel venture]

yo mr. pancake imma go but hope you get everything sorted out gn gang

[analog relic]

But just keep practicing

[steel venture]

also @analog relic that pfp is unsettling

[analog relic]

pfp?

Oh

Ya

I should change it

[steel venture]

as is the name but that's just cause i don't know

russian?

arabic?

ukranian?

some eastern european country's idiom?

[novel dirge]

doesn't look eastern european

maybe just greek symbols

[steel venture]

oh

yeah

alpga

huh

[novel dirge]

I still get the same result

damn

[harsh cipher]

Hi

y= -5log_(x-3)+2

how do I determine the equation of the asymptote when there is log infront of (x-3)

I considered inputting the values for x but could not solve it correctly

[slender bay]

Take y=logx

Transform to y=-5logx

Then do y=-5logx+2

Then do y=-5log(x-3) +2

[willow bear]

y= -5log_(x-3)+2

are you missing the base there

[slender bay]

10

Or e

[willow bear]

i'm not asking you

[slender bay]

Its obvious..

[harsh cipher]

yes it's base 10

I learned that in the previous lessons

[slender bay]

So if the processes make sense..u will have ur ans.

[harsh cipher]

it did not make sense tbh

[slender bay]

Oh....if u know the curve of y=f(x) can u draw y=-f(x)

[harsh cipher]

reflect on x axis

[slender bay]

Yeah..

So if u mumtiphy with 5 the later curve

How will it look like

[unique hill]

the 5 just makes it narrower

[slender bay]

It streches the curve along y axis

[harsh cipher]

sure but how is this relevant to getting the asymptote?

lol

[slender bay]

Wr are getting there

[unique hill]

5 doesnt change the asymptote

[slender bay]

Have patience

[harsh cipher]

okay

[slender bay]

Then add a 2 to the streched curve

The y gets above by two units

So we have got to y=-5log(x)+2

Right

[harsh cipher]

hmmm

[slender bay]

Now for y=-5log(x-3)+2 ....just shift y=-5logx+2 by 3 units towards right purely in the x direction

Now can u see where is the new position of the asymptote

[harsh cipher]

i got that part but didn't get the first explanation....

[slender bay]

The 0 point of y=-5log(x-3)+2 is at x=3

Look at the inside

x-3

[harsh cipher]

yea

[slender bay]

So u understood ?

The new position of asynptote is x=3

Which was previously at x=0

[unique hill]

nice step-by-step expln

[harsh cipher]

I did not understand still

I still did not understand...

[unique hill]

asymptote shifts along with the graph

[slender bay]

Then wht more can i do

[harsh cipher]

nothing

[slender bay]

Ask somebody else

[harsh cipher]

thank you anyways

[slender bay]

Yeah. ..wasted

[harsh cipher]

no I'm reading over what you wrote

[unique hill]

which bit u dont understand?

[harsh cipher]

https://discordapp.com/channels/268882317391429632/363224154469826562/694110801082056815

Take y=logx
Transform to y=-5logx
Then do y=-5logx+2
Then do y=-5log(x-3) +2

this part

lolll

[unique hill]

the (x-3)?

[slender bay]

I can introduce the techniqalities of an asymptote...but it will be a mess then..i leave it to nighty

[unique hill]

(x-a) shifts a units to the right

[harsh cipher]

I know that

[unique hill]

so which bit u dont understand

[harsh cipher]

I don't understand why the asymptote is at x= 3

I think it should be at x = 2?

[unique hill]

before u shift the graph 3 units to the right, the asymptote is at x=0

[harsh cipher]

yes

[unique hill]

and after u shift the graph, the asymptote also shifts itself 3 units to the right

so x=0 --> x=3

[harsh cipher]

ok thank you I need to look at my notes where is said it the asymptotes shift

[unique hill]

lmk if u need help

[harsh cipher]

ok ty!

[unique hill]

always happy to help

[leaden stratus]

@lilac pier I found the way to solve it

[lilac pier]

nice

how

[leaden stratus]

how
@lilac pier tan a/2 is also 1-cosa/sina, so you get this...

[lilac pier]

anyways you got it, thats what matters

good job

[leaden stratus]

@lilac pier so, this was really easy. I wondered how you could do it using tan a/2 = sqrt(1-cos/1+cos)

[lilac pier]

use conjugate

[maiden igloo]

I am having a little trouble with this question

I am being asked to find the domain of this function

h(x) = 3-2log(x/2-5)

A little confused and not sure where to begin

[sharp marsh]

How do I do this?

e^2x − 9e^x + 20 = 0

[maiden igloo]

pretty sure you can use log to cancel out e

if i remember correctly

[hasty heath]

@sharp marsh substitute lets say k for e^x, and your formula would be

k^2-9k+20 = 0

once you solve for k

[maiden igloo]

thing i found my answer

x > 10

[hasty heath]

oh subi

ngl if u wanna quickly find answer

just graph it on desmos

and look lol

[maiden igloo]

i am trying to learn

haha

[hasty heath]

use that to check it

thats what id do

[maiden igloo]

so i try to save that to check

yeah

[hasty heath]

yea

[maiden igloo]

that's what i did

[hasty heath]

well for ur functiont

the domain right?

[maiden igloo]

yeah

it would just be

x/2 -5 > 0

[hasty heath]

yeah

[maiden igloo]

• 5 both sides

awesome

[hasty heath]

jsut find where x can't be some value

and thats only true for log(x) in that formula

so just solve that

[maiden igloo]

sometimes i trip out when new things are added

[hasty heath]

u got it

lol yaeh

just looka t it by parts

[maiden igloo]

yeah learning that

appreciate the help

[sharp marsh]

@hasty heath k^2-9k+20?

[hasty heath]

yeah

replace e^x with k

like substitute it

[sharp marsh]

um

What happens to e then?

[hasty heath]

e^x becomes k

so you factor out e^x, and replace it with k

and onec you solve for k

you just do

k = e^x and solve for x then

[sharp marsh]

Wait I get it now

But like why are you allowed to replace e^x with k?

or any other variable?

[steel tulip]

That's just something you can do... letting a variable equal something else

Since it doesn't change the original equation

[hasty heath]

@sharp marsh well if I do e^x = k

that doesn't actually change the formula at all like tiessie was saying

cause anywhere we use k, we can just substitute (e^x) back in

can't we?

a simpler example would be like

4x + 12 = 8

substitute 4 with m

so now we have 2 formulas

4 = m

and mx + 3m + 2m

the formula is different yeah, but we just replace 4 with m

[steel tulip]

and mx + 3m + 2m
@hasty heath mx + 3m = 2m

[hasty heath]

oopsie lol ty

[sharp marsh]

oh

log9x -log9(x-8) = 1/2
You get log9(x/x-8)=1/2 right?

[barren oasis]

Tried this for two or three hours now and I can't figure it out. I did every other problem on this worksheet NO problem... any help is appreciated. I tried starting right side and breaking down everything into sines and cosines and then tried reducing the fraction but I just got sin(x) over 1- (cos(x))^2

[fleet yew]

Ok

@barren oasis 1 - cos^2 = sin^2

[barren oasis]

Yeah, I got that too, but how do I go from there to get sec(x)/tan(x) ? We're trying to verify this identity Im sorry

[fleet yew]

Honestly i would restart the problem

Cross multiply

Tan^2=sec^2-1

Which is true

[barren oasis]

I would do that but he said we can't, we're not allowed to do that

we're not allowed to treat this like an actual algebra problem, we have to start with one side and go from there until we get the other

[fleet yew]

Lets start with the left side then

sec/tan = (1/cos)/(sin/cos)

Multiply top and bottom by cos

You get 1/sin which is csc

[barren oasis]

But don't we need secx/tanx not cscx? and I don't think it would be helpful to use the cotangent identity for that

[sharp marsh]

Use a graphing utility to graph the exponential function.
y = 4^(x + 1) – 3

[barren oasis]

If you don't have a graphing calc just use desmos, it's a really good tool

[sharp marsh]

Oh

Yeah I forgot about that]

I'm on a different problem rn though

y = −log5 x + 1

[barren oasis]

what are you trying to do?

[sharp marsh]

I have to get the domain, vertical asymptote, and x intercept

[barren oasis]

are you allowed to use the graph? if you can then that's all very easy to see.

[sharp marsh]

Nope

[barren oasis]

if not then you'll need to do some testing on which values are or aren't holes and what is or isn't included

[sharp marsh]

I just have to figure it out

[barren oasis]

for the domain you can just solve where you know a logarithmic equation isn't in the domain (because a logarithm is an exponent) and for the asymptotes youll have to just try some numbers, and x intercept should be a given, where does your x = 0

[sharp marsh]

um

It's negative log though

what 2 do

[barren oasis]

yeah it's still a logarithm it just has a reflection now

make any sense??? your vertical asymptote should be (idk if it is for sure) just the y axis, yeah? and if your asymptote is at the y axis that gives me a hunch my domain excludes the negative numbers

[sharp marsh]

um

[barren oasis]

what's wrong?

[sharp marsh]

I don't get it

How do I get my domain then

[barren oasis]

your domain is just the set of x-values which exist

okay? so if you take a few numbers you can get a good idea where exactly your boundaries are

i'd just use 0, 10, and -10, if you dont wanna think about it much, and just solve, if you get something undefined, that's not in your domain

and your vertical asymptote is where your function acts up, because it's approaching a value which isn't in the domain, okay?

If anyone could provide a little more insight to me that'd be helpful... Im having some issues verifying... I've worked this out a bunch of times and I swear im just missing out on something really obvious I should see -- Ive been working on the right side trying to use complex fractions to reduce but I keep getting something that has a secant or a tangent but not a secant over a tangent (verifying this identity)

[hardy abyss]

@barren oasis can you compute (sec x) * (sec x - cos x) and see if you get anything useful?

[barren oasis]

I tried doing the conjugate of it to combat this problem, but whenever I do that, I get a "-2secxcosx" term in the middle and I don't think that is going to help any, would it??

[hardy abyss]

is that a reply to my suggestion?

you shouldn't need to do anything besides multiply to evaluate the expression I posted

[barren oasis]

sorry I had to go do something and well I dont know what I'd do after that is all, you'd be left with secants and cosines, right?

[hardy abyss]

well, I would suggest that instead of speculating about what you would get, to just do it, it only takes one step
but I scrolled up and saw that you already asked this question and rejected the path I was going to lead you on

so in anticipation of you not liking this solution, I'll propose that we take another approach, capiche?

[barren oasis]

Any solution works, I just went with my teacher's suggestion, he said to turn everything into sines and cosines etc etc but I don't know if I did it right

[hardy abyss]

well someone already gave you the solution I gave you and you said that you "can't do that"

so let's turn everything into sines and cosines

what do you get when you turn (tan x) / (sec x - cos x) into sines and cosines?

[barren oasis]

you get sin(x)/cos(x) / ((1/cos(x) - cos(x))

[hardy abyss]

okay, and that is...?

[barren oasis]

sin(x)/cos(x) / ((1 - cos^2(x)))/cos(x)

[hardy abyss]

ok, that works

[barren oasis]

then I multiplied by cos(x) to get rid of the complex fraction? and I got sin(x) over the 1 - cos squared bit

[hardy abyss]

yup

[barren oasis]

then I got stuck and I kept trying to think I had to do something with the pythagorean trig identity?

[hardy abyss]

yes that is exactly what you should use next

[barren oasis]

I did that, but then I just got sin(x)/sin^2(x)

[hardy abyss]

yep

[barren oasis]

but how does that help? I had no idea what to do after that because getting rid of the sine in the numerator would just give away my tangent

and I would need a cosine in the denominator but if I try to put a cosine in the denominator I end up putting it in the numerator

[hardy abyss]

so at the end of the day you got that the RHS is equal to 1/sin(x) yes?

[barren oasis]

Yeah... but my RHS doesn't equal my LHS?

[hardy abyss]

what is sec(x)/tan(x) in terms of sine and cosine?

[barren oasis]

1/cos(x) / (sin(x)/cos(x))

[fleet yew]

Parantheses

[hardy abyss]

go on

[barren oasis]

oh sorry

which is just 1/sin(x)

[hardy abyss]

right

[fleet yew]

1/sinx = 1/sinx

Can you do anything else to that equation

[hardy abyss]

very enlightening thank you

[barren oasis]

well, no, but I have to make it sec(x)/tan(x)

my teacher doesn't want us to break down both sides and do that

he wants us to pick and choose a single side and that's where I get stuck

[hardy abyss]

there is no reason that you can't write everything you just did as one long equality

[fleet yew]

Well you got that the RHS = 1/sin

So multiply top and bottom by sec

Then you have

Sec/tan = sec/tan

[barren oasis]

on a quiz or a test though I would never be able to think about that without breaking it all down

[fleet yew]

A good way is to think about it backwards

Once you boiled the right side down to 1/sin

You need to multiply it by an identity to get sec/tan

In other words

1/sin * a/a = sec/tan

Where a is the number that will make that statement true

In this case a = sec(x)

So it works

[barren oasis]

Oh okay, that makes sense. Thank you. I feel dumb now that I didn't really notice that

[fleet yew]

👍🏻

@barren oasis you look like a kid from my AP gov class

[steel venture]

where are you stuck

[deft narwhal]

i dont understand it 😦

[steel venture]

ok no problem

so theyre asking you to find x correct?

and in order to do so it would be helpful to isolate the x

can you do that or do you want me to walk you throuhg it?

[deft narwhal]

so i subtract 1 and get 6sin(20x)=0

then i divide 6sin and get 20x=(0/6sin)??

[steel venture]

nono

so sin is a trigonometry function

so first thing you can do is divide by six

which gives you sin(20x) = 0

and now we do what is called the inverse sin

are you following

it would look something like this:
sin^-1(sin(20x)) = sin^-1(0)

[sharp marsh]

I still don't know how to get the domain for this

y = −log5 x + 1

[steel venture]

what are the possible x values that give a real y? @sharp marsh

[sharp marsh]

Um

Doesn't it have to be positive?

[steel venture]

yes it does have to be positive!

[sharp marsh]

Wait how do I get domain then

Do I not have to do the math because it's not in parenthesis?

[steel venture]

what do you mean?

[sharp marsh]

Um idk

Wait how do I get domain

I'm lost

[steel venture]

domain is the x values that make the function work

so for y = x the domain is: x can be any real number

for 1/x, x can't be 0

[sharp marsh]

Doesn't x always have to be positive?

[steel venture]

no

[sharp marsh]

um

[steel venture]

when you have log(x)

the x does have to be positive

but in the case of y = x, or y = x^2, x doesn't necessarily have to be positive

[sharp marsh]

oh

So what do I do to get domain

since log is negative

[steel venture]

so the actual y value that we get is not too important

what matters is what we put inside the log

[sharp marsh]

I'm lost now

[steel venture]

so lets take y = log(x)

when we're talking about the domain, we want to know what the possible values of x arre

do you get that?

[sharp marsh]

ya

[steel venture]

so in order to find what works and what doesn't, we need to know when log(x) works and when it doesn't

so log(x) is only defined when x is greater than 0

log(-1), log(-pi), log(-23412431234) are all undefined

this is because the log function needs to take a positive input

does that make sense

[sharp marsh]

ya

[stuck lark]

make branch cut, log(-1)=i*pi

[sharp marsh]

uh

[steel venture]

shhhh buncho

stop confusing people

[stuck lark]

call me rocket/rokabe

[steel venture]

don't worry he's right, but its beyond what we're doing

[stuck lark]

buncho is a nickname fad

[steel venture]

ok mb

[sharp marsh]

[steel venture]

yeah i've noticed

@sharp marsh so in order to find what works and what doesn't, we need to know when log(x) works and when it doesn't
so log(x) is only defined when x is greater than 0
log(-1), log(-pi), log(-23412431234) are all undefined
this is because the log function needs to take a positive input
does that make sense

just worry about what i said here

so the domain is when the x makes the funciton work

[sharp marsh]

ok

[steel venture]

so now

if i gave you log(x) what would the domain be?

[sharp marsh]

(0,infinity)

[steel venture]

good!

what about log(x-1)?

[sharp marsh]

(1,infinity)?

[steel venture]

exactly

what about log(-x)

[sharp marsh]

uh

(-infinity,0)?

[steel venture]

yep

ok

so i think you get the gist of what domain means now

[sharp marsh]

but like

in
y = −log5 x + 1

[steel venture]

y = −log5 x + 1

yeah ok

[sharp marsh]

is x+1 suppose to be in parenthesis

[steel venture]

is it?

[sharp marsh]

or would it only be the x?

uh

no

?

[steel venture]

do you have the question itself

[sharp marsh]

Find the domain of the logarithmic function. (Enter your answer using interval notation.)
y = −log5 x + 1

[keen vapor]

How do i solve 64^5x+2=8 with no calculator

[steel venture]

ok useless we can do both real quick

so lets start with the easy one

y = −log(5 x) + 1

now

what needs to be true for the function to work?

[sharp marsh]

x has to be negative right?

[steel venture]

well lets go back a few steps

the domain for log(x) was (0, infinity) right?

so what is the domain for -log(x)

[sharp marsh]

(-infinity,0)

?

[stuck lark]

the natural domain of $x\mapsto-\log(x)$ isn't $(-\infty,0)$

[obsidian monolith]

RokettoJanpu:

[sharp marsh]

uh

[steel venture]

so remember

this is the same thing as -1 * log(x)

nolan

[keen vapor]

Yo

[steel venture]

give me a sec bro

[keen vapor]

ur good

[sharp marsh]

(1,infinity)?

i'm confused

[steel venture]

yeah so lets take it abck a notch

log(x)

domain is (0, infinity)

what about 2log(x)?

[sharp marsh]

uhh

idk

[steel venture]

ok

what does domain mean again?

[sharp marsh]

number that x needs to make y positive?

[steel venture]

nope

its an x value that makes the function work

so if we have y = log(x)

log(x) only works when the stuff inside the log is greater than 0

[sharp marsh]

ya

[steel venture]

so that means if we have 2*log(x)

has that changed?

wait dont answer that

for log(x) to work

x>0 right?

[sharp marsh]

yes

[steel venture]

for log(z) to work

z>0 right?

[sharp marsh]

,yes

[steel venture]

what about for log(w)

what needs to be true

[sharp marsh]

w >0

[steel venture]

perfect

so for log(2x) what needs to be true

[sharp marsh]

x > 1/2?

[steel venture]

ok lets do this first

imagine 2x = m

m is just a random letter

now

we can do log(2x) = log(m) right?

[sharp marsh]

ya

[steel venture]

so that means that now we have log(m)

so what needs to be true

[sharp marsh]

wym

[steel venture]

for log(m) to work

[sharp marsh]

u have to log m?

[steel venture]

so for log(x) to work

what was the case that needed to be true

[sharp marsh]

isnt it log 2x?

or just x

[steel venture]

just think about the x scenario

if we have log(x)

what needs to be true for the function to work

[sharp marsh]

x > 0

[steel venture]

perfect

so what about log(m)

[sharp marsh]

m > 0

[steel venture]

so

we know that m = 2x right?

[sharp marsh]

oh

ya

[steel venture]

so what does m > 0 look like

when we switch m for 2x

[sharp marsh]

m > 1/2x?

[steel venture]

nono

m > 0

[sharp marsh]

uhh

[steel venture]

here

m > 0 right?

[sharp marsh]

yes

[steel venture]

now lets say m = n

so then if m > 0

what can we say about n

[sharp marsh]

n /

n > 0

[steel venture]

ok

so now back to log(2x)

if 2x = m

then log(2x) = log(m) right?

[sharp marsh]

um ya

[steel venture]

ok

so if we have that for log(m) to work

we need m > 0

and m = 2x

then what can we say about 2x

[sharp marsh]

2x > 0

[steel venture]

perfect

so what does that tell us about x

(hint divide both sides by 2)

[sharp marsh]

x > 0?um9, ubdubuty

[steel venture]

yeah

so now for log(2x)

what is the domain

[sharp marsh]

uh

2x >0?

[steel venture]

yes

and what does that mean about x

[sharp marsh]

x isc also > 0?

[steel venture]

yep

so what is the domain of x

[sharp marsh]

0,infinity

[steel venture]

perfec

t

so

what about log(3x)

[sharp marsh]

3x > 0

[steel venture]

so what is the domain

[sharp marsh]

0.infinity

,

[steel venture]

perrfect

now what about log(1351.75442x)

[sharp marsh]

0.infinity
@sharp marsh

[steel venture]

ok

good

now what about 2log(x)

[sharp marsh]

0.infinity

[steel venture]

perfect

so what about 6log(x)

[sharp marsh]

0.infinity

[steel venture]

what about a * log(x)

a can be any number

[sharp marsh]

0.infinity

?

[steel venture]

yep

so what if we set a to -1?

-1 * log(x)

or -log(x)

what is the domain?

[sharp marsh]

0.infinity0.infinity0.infinity0.infinity0.infinity0.infinity0.infinity0.infinity0.infinity0.infinity

[steel venture]

exactly

so not

now

y = −log5 x + 1

y = −log(5x) + 1

what is the domain of −log(5x)

[sharp marsh]

0.infinity

?

[steel venture]

yep

what about −log(5x) + 1

[sharp marsh]

same thing?

[steel venture]

yes

what about −log(5x + 1)

[sharp marsh]

-infinity,1/5?

[steel venture]

nono

do this first

[sharp marsh]

arme

[steel venture]

log(x + 1)

[sharp marsh]

aren't u doing 5x + 1 = 0?

oh

thats just -1,infinity

right?

[steel venture]

yep

so

what about log(5x + 1)

[sharp marsh]

-1?, infinity?

[steel venture]

so lets work it out

5x + 1 > 0 right

[sharp marsh]

mhmm

[steel venture]

ok

so keep going

show me your steps

[sharp marsh]

5x > -1

x > -1/5

[steel venture]

perfect

you had it haha

[sharp marsh]

wait

Don't I need to flip the sign?

Cause i'm dividing?

The > over

[steel venture]

thats only when you divide by a negative number

[sharp marsh]

OH

[steel venture]

so ok

what about -log(x+1)

[sharp marsh]

-1,infinity

[steel venture]

and what about -log(5x+1)

[sharp marsh]

1/5, inf

[steel venture]

wait so lets do that again

what has to be true again

[sharp marsh]

5x + 1 > 0?

[steel venture]

yep

[sharp marsh]

-1/5, inf

[steel venture]

there you go

so thats the domain

but since its > and not >= then it has to be exclusive

i.e. x cannot be -1/5

[sharp marsh]

but in y = −log5 x + 1

What is in parenthesis then?

[steel venture]

so thats the thing

idk if its y = −log(5 x + 1)

or

y = −log(5 x) + 1

[sour inlet]

Having trouble graphing f(x) = tan (2x - pi) + 1

So far I know my amplitude is 1
My period is pi/2
my x-scale is pi/4
and my phase shift is pi/2
and my vertical shift is one.

This is what I have so far but I just don't know where to go from here and what needs to be graphed

[steel venture]

OH THATS LOG BASE 5 lmfao

[sharp marsh]

so -log5 (x+1)?

o

[steel venture]

i think its just log_5(x) + 1

so what is the domain?

[sharp marsh]

0,infinity?

[steel venture]

yep

pomf graph some simple points

[stuck lark]

i dislike when hw sites fail to use brackets when needed

[steel venture]

start by doing x = 0

x = pi/2

x = pi

x = 2po

pi*

[sour inlet]

Are we just inputting random numbers to see what comes out? I'm confused.

[steel venture]

yes

so you can get a general shape for the graph

and from there fill in the pattern and curve for the whole graph

[harsh cipher]

question

Hello

[steel venture]

hello

[harsh cipher]

Hi

question b

[sour inlet]

x = 0 , gives tan(-pie) which is 0
x= pi/2, gives tan(pie) which is 0
x = pi, gives tan(pie) which is 0
x = 2pi, gives tan(3pie) which is 0

I'm confused, all the outputs end up being 0

[steel venture]

mirrion:
log(x^a) = alog(x)
log(a/b) = log(a) - log(b)

log(ab) = log(a) + log(b)

[sharp marsh]

y = −log5 (x) + 1

How do I find vertical asymtope then?

[steel venture]

pomf plot them all as zero then

and plug in other points

like pi/4

pi/6

pi/12

and other unit circle points

[harsh cipher]

I understand the log laws you explained

[steel venture]

@obsidian path as x-> 0 , what does y -> ?

[harsh cipher]

but it says that log_a(x) = 4

[steel venture]

@sharp marsh nvm that was wrong

[sharp marsh]

lol

[steel venture]

@obsidian path sorry didnt mean to

[sharp marsh]

Just searched it up

It's just = sign

instead of > right?

[steel venture]

nono i meant what is the smallest x value possible

like remember the domain

what is the value it could never reach

mirrion i don't see the issue

[harsh cipher]

we have denom log_a(ay^2)

where is the x ?

lol

[steel venture]

log(a/b) = log(a) - log(b)

[harsh cipher]

therefore

[steel venture]

so then you have log(x^(1/2)) - [log(ay^2)]

log(ab) = log(a) + log(b)

log(x^(1/2)) - [log(a) + log(y^2)]

all the bases are a btw

just dont feel like writing it out

[harsh cipher]

so what happens to this part [log_a(a) ?

[steel venture]

a raised to what power = a?

[harsh cipher]

yea

because log_a(x) = 4

I guess 1?

[steel venture]

yes

[harsh cipher]

very tricky question

got it

very easy now

@steel venture thanks!

[steel venture]

😄

[maiden igloo]

Working with exponential equations a little stuck on this

8 = 4x^2 · 2^5x

Watched a video but it had common bases

[steel venture]

looks like a calculator equation

[maiden igloo]

uhh don't wanna do that

wanna learn 🙂

[steel venture]

there's no simple way to solve this i don't think

[daring yarrow]

cant u put everything on base 2? or am i missing something

[steel venture]

as you have a variable in the exponent as well as the base

wait

is it 8 = 4x^2 · 2^(5x)

or

8 = 4x^2 · 2^(5)x

[maiden igloo]

oh rip

my b

8 = 4^x · 2^5x

[fleet yew]

Ok

4= 2^2

4^x= (2^2)^x

= 2^(2x)

[daring yarrow]

4^x? wheres it say that

[maiden igloo]

hold

let send pic of where I am at

to make sure I am understanding

I am sorry

correct equations was

8 = 4x^2 · 2^(5)x

[daring yarrow]

ah

[fleet yew]

2^(5x), or 2^5 * x ?

[steel venture]

alright that looks better

now look at the 8 and the 4

[maiden igloo]

Yes sir

[steel venture]

they are both numbers that cna be expressed as a power of 2

[maiden igloo]

i can do

8 = 2^3

and 4 2^2 ?

[steel venture]

yep