help-10

[fervent tusk]

but it seems you understand the basics

just check the domain part of the function and substitute it appropriately

[timid silo]

thank you

also

how do i do these bottom two?

[fervent tusk]

im so confused

is that graph your own

[timid silo]

no

basically for the ones i did

you start at the point

and basically draw a line to an endpoint

[fervent tusk]

yeah that makes sense

but the last two are functions

and they asked you to evaluate them

there could be mistake in this question

[timid silo]

theres 2 other questions exactly like these

that i have

[fervent tusk]

send them

[fervent tusk]

0x is 0 which is the x-axis

it doesnt make any sense

sorry mate

i don't think i can help you with this

[timid silo]

its alg

[obtuse pebble]

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[obtuse pebble]

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[silk tiger]

how does it define a polynomial function on the parabola

[meager trench]

sqrt(x) = y

y is a polynomial on X

[obtuse pebble]

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[tardy epoch]

Show how you got your numbers

[obtuse pebble]

@cedar talon Has your question been resolved?

[obtuse pebble]

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[storm sluice]

How do I solved for a?

Is a just pi?

[thin star]

I thought someone was helping you

[storm sluice]

Yeah, but it was like 20 minutes in between convo so I thought I’d restart

@fervent cradle

<@&286206848099549185>

[obtuse pebble]

@storm sluice Has your question been resolved?

[fervent cradle]

yeah it's just pi

as h goes to zero the thing in the top just becomes sin(pi)

so it's like at pi

[obtuse pebble]

@storm sluice Has your question been resolved?

[storm sluice]

but 0 is in the denominator?\

@fervent cradle

<@&286206848099549185>

[fervent cradle]

right

it's a derivative

as h goes to 0 the top becomes 0 also

so you get 0/0 sorta which is fine

comparatively

[storm sluice]

close

.close

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[tame zinc]

honestly don't understand why i'm wrong here

[tame zinc]

aren't all linear function's domains -inf to inf

just type directly in channels under math help

<@&286206848099549185>

.close

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[tribal scaffold]

i dontt really get this one, tthe answer said a = b = c = 0 and d = k, but it seems wrong because that way it will overlap and will meet more than 4 times

[royal basin]

they didn't say exactly four times

which is a bit of an um-ackshually but is also the only interpretation that makes sense

[tribal scaffold]

ohh ok

so if it doesnt say exactly it means its 4 or more?

alr tysm

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[junior hearth]

how do i start on number 35, do i graph it? and how

[fossil crag]

Do you know what $\lim_{x\to (\pi/2)^+}\cos(x) = ?$

[warm shale]

rafilou2003

[junior hearth]

no i do not

[fossil crag]

What is cos(pi/2)?

[junior hearth]

would I have to look at the graph of cos x and find the limit?

[junior hearth]

i forgot lol

[fossil crag]

You can do that if you want to

[junior hearth]

cos pi/2 is 0 right

[fossil crag]

Yes

And after that is it positive or negative?

[junior hearth]

what do u mean?

[fossil crag]

Right after x>pi/2, is cos(x) positive or negative?

[junior hearth]

i'm not sure what you're asking

[fossil crag]

Look at the graph for cos(x)

Look immediately to the right of the point pi/2

[junior hearth]

ohh its positive

wait no

negative

[fossil crag]

Yes negative

So when we approach it from the right, $\lim_{x\to (\pi/2)^+}\cos(x) = ?$

[warm shale]

rafilou2003

[junior hearth]

is 0?

[fossil crag]

0 yes, but more precisely if possible ?

0+ or 0-?

[junior hearth]

i don't know

what 0+ or 0- means

[fossil crag]

Ok, well shouldn't be too hard to do without those

Can you apply what we just did to your limit that you want to find?

[junior hearth]

i'm assuming you want me to look at the graph of 1/x secx right

[fossil crag]

Not necessarily

[junior hearth]

and to see if approaching pi/2 from the right is -infinity or positive

i know that sec is 1/cosx but idk how that leads to what we talked about

[fossil crag]

Well you can try without the graph

So, plugging in pi/2, you know the limit is gonna be infinite

What remains is to know if it's positive or negative

[junior hearth]

what are we plugging pi/2 in to specifically

sorry my professor didnt go over that

[fossil crag]

1/x sec(x)

[junior hearth]

so it would be 1/ pi/2 sec (pi/2) ?

[fossil crag]

Yes, and sec(pi/2) in this particular case is?

[junior hearth]

let me take a look at the graph

well in the graph of secant it says pi/2 is an asymptote

[fossil crag]

Yes

Because cos(pi/2) = 0

[fossil crag]

And so when you approach it from the right side, do you get + or - infinity?

[junior hearth]

negative infinity

[fossil crag]

Yes

[junior hearth]

wait

i'm confused we just looked at the graph of secant

but in the problem it is 1/x sec(x)

what do we do with 1/x?

[fossil crag]

What does 1/x approach when x-> pi/2+ ?

[junior hearth]

i'm not sure how to solve that

when x is approaching pi/2 from the right side it is negative infinity

but idk how to solve for 1/x

[fossil crag]

Well it shouldn't be a problem should it?

1/x is continuous over the set of positive numbers

So 1/x goes to 1/(pi/2)

[junior hearth]

and what do we do with 1/(pi/2)?

[fossil crag]

Well It's a positive constant

And what can we say about positive constant × negative infinity ?

[junior hearth]

it is negative infinity

[fossil crag]

Yep

[junior hearth]

ohhhhh

wait so can u check if i did my problem right

i will show the work out

is that how you’re supposed to show the work?

[fossil crag]

I don’t know how rigorous your teacher is but that's the gist of it!

[junior hearth]

so you can solve this problem by also looking at the graph of 1/x sec(x)?

[fossil crag]

You could also do that

Also I suggest you write sec((pi/2)+) instead of sec(pi/2)+

[junior hearth]

why is that

oh

[fossil crag]

Because sec(pi/2) is 1/0

So if you take sec(pi/2)+ that could be considered as 1/0+

(Which is + infinity)

[junior hearth]

i see

can you also help me on one more question

[fossil crag]

Sure

[junior hearth]

its asking to determine the infinite limit

but how do i approach this?

[fossil crag]

I suggest factoring the denominator

[junior hearth]

ok so after that i get (x-1)(x-1)

what do i do next im lost

[fossil crag]

Well it's (x-1)²

What does this approach when x-> 1?

[warm shale]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[junior hearth]

do i do this by plugging in 0.9 for x

[fossil crag]

Just plug in 1

[junior hearth]

well its 0

[fossil crag]

Well yep

And the numerator ?

[junior hearth]

3

[fossil crag]

Yes

[junior hearth]

3/0 is undefined tho

[fossil crag]

Yes but here it's not just 3/0

We know that (x-1)² is always...

[junior hearth]

idk

[warm shale]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[fossil crag]

Well it's a number squared

So it's either 0 or positive

[junior hearth]

oh right

[fossil crag]

So it's 3/0+

[junior hearth]

what does this mean for the problem tho

[fossil crag]

Well that should give you the limit no?

3/(0+)

[junior hearth]

idk how to find the limit with that info

[fossil crag]

Ok, what happens when you divide 3 by an infinitely small number?

[junior hearth]

ohhhhh

positive infinity

[fossil crag]

Yes !

[junior hearth]

oh my god thx bro

[fossil crag]

It's important to remember that 1/(0+) = + infinity
And 1/(0-) = - infinity

[junior hearth]

so does that mean the limit is DNE?

since they are different

[fossil crag]

Not here

Because we know that (x-1)² has limit 0+ ALWAYS

[junior hearth]

but i thought if the limits differ then it is DNE

[fossil crag]

Whether from the left or the right, It's always positive before reaching 0

[fossil crag]

Yes this is true

But here the limits don’t differ

It's always 3/0+

[junior hearth]

i don't understand but maybe i will learn this later

so when finding the infinite limit should i always start by plugging in the value?

that x approaches

[fossil crag]

Yes It's a good reflex, even if it doesn't give anything

If you have something of the form a/0 then ask yourself whether the 0 is a 0+ or a 0-

If it's neither then it's DNE

[junior hearth]

how would i know if it is 0+ or 0-?

in the case that theres no square

[fossil crag]

We can do what we did for example with sec

We checked that cos(x) was negative on the right of pi/2

So it was a 0-

Since you won't always have access to all the curves you want, it's nice to start doing those types of reasoning

[obtuse pebble]

@junior hearth Has your question been resolved?

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[tame zinc]

can someone expalin to me what is wrong here

comfused on the h(x) values

[frank monolith]

h(x) = g(f(x))

lemme give you an example

let f(4) = 2 and g(2) = 9

so h(4) would be g(f(4)) = g(2) = 9

[obtuse pebble]

@tame zinc Has your question been resolved?

[bronze dirge]

. reopen
If someone can answer my doubt please help me out 👍

[obtuse pebble]

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[tame vale]

How do we get $\log_{2}({\sqrt{6}+2})$ from the equation $2^x-2=(\frac{1}{2})^{x-1}+2$?

[warm shale]

Bennxy

[rich plume]

wdym get?

like you need to solve for x?

[tame vale]

Yes

[rich plume]

show your work

[tame vale]

Ok

Wait a moment

Please ignore those equal and greater than sign

I think I messed something up at step 5

[obtuse pebble]

@tame vale Has your question been resolved?

[tame vale]

<@&286206848099549185> Can someone see what went wrong for the equation 2^x-2=(1/2)^{x-1}+2?

[frank monolith]

how did an inequality change to an equality to an inequality?

[thorny quail]

is it greater than equal to 3/4? the answer

[tame vale]

Ehh yea I mentioned that earlier, but was automatically writing it as equality. And didn't change it afterwards

[tame vale]

No it was $\log_{2}({\sqrt{6}+2})$

[warm shale]

Bennxy

[frank monolith]

3rd last step is wrong

also, confirm this, equality or inequality?

[tame vale]

The original question was an inequality

[tame vale]

Why?

[warm shale]

Dyssrupt

[tame vale]

Hmm okok

But how do we proceed now

We can't simplified or split log_{2}(2^{-x}*2+2^{2})

[frank monolith]

let 2^x = t

and form a quadratic in t

[tame vale]

So t=2*t^{-1}+4?

And then 2*t^{-1}-t+4=0?

[frank monolith]

its better to write it on paper

and send

[tame vale]

I don't have any paper with me right now, so I write latex to show you.

So when I use 2^x=t

After using the quadratic formula we will get

$2-\sqrt{6} \vee \sqrt{6}+2$

[warm shale]

Bennxy

[tame vale]

Because t=2^x

We get $2^x=\sqrt{6}+2$

[warm shale]

Bennxy

[tame vale]

So that means x=$\log_{2}({\sqrt{6}+2})$

[warm shale]

Bennxy

[tame vale]

This calculating is correct right <@&286206848099549185>?

[frank monolith]

yes, it is

[tame vale]

Aha nice

Thanks for the help man

.close

[obtuse pebble]

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[tidal nymph]

hello, i am using geogebra and i need to create two triangles with two equal sidses and one equal angle, but they cant be congruent, so im trying to find a way so that moving the point B will automatically move point D so that CB and DC are the same distance

[obtuse pebble]

@tidal nymph Has your question been resolved?

[tidal nymph]

found a way

[obtuse pebble]

@tidal nymph Has your question been resolved?

[obtuse pebble]

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[coarse hound]

to do this you have to multiply divide and subtract things from the right hand side until youre left with only v on it

[lone echo]

you dont do things to one side.
You do the same to both sides

[coarse hound]

sorry i wasnt clear i mean to do those operations to both sides until youre left with v while focusing on clearing up the right hand side

[reef hull]

thank you guys

.close

[obtuse pebble]

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[pearl knoll]

can someone help me simplify this question:

[pearl knoll]

go step by step please

[spring steeple]

Simplify the numerator first

[pearl knoll]

yes but how

i know i need a common denominator

i just dont know how to do that

[spring steeple]

Make a common denominator then

[pearl knoll]

i dont know what it is

[spring steeple]

Make $\sqrt{x}\sqrt{x+h}$ the common denominator of the numerator

[warm shale]

Normed

[pearl knoll]

ok

[spring steeple]

Once you are done doing that multiply both numerator and denominator (of the numerator of the original expression) by the conjugate of the numerator (of the numerator of the original expression)

[pearl knoll]

???

[spring steeple]

You know conjugate?

[pearl knoll]

wouldnt i multiply 1 over square root x+h both numerator and denominator times square root of x+h times square root of x?

and then do the same for the 1 over square root of x?

[spring steeple]

Ahh wait let's go step by step

[spring steeple]

Have you done this?

[pearl knoll]

yes

[spring steeple]

got something like $\frac{\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x} \sqrt{x+h}}}{h}$?

[warm shale]

Normed

[pearl knoll]

no

[spring steeple]

What then?

[pearl knoll]

im confused on the numerator

[spring steeple]

Alright forget the h for a moment and just focus on the numerator, so when you made $\sqrt{x}\sqrt{x+h}$ the common denominator, what you got on the numerator?

[warm shale]

Normed

[pearl knoll]

for 1 over x+h(square root) would you multiply top and bottom by square root of x+h and square root of x?

[spring steeple]

No just multiply top and bottom by sqrt(x)

[pearl knoll]

oh ok

whats 1 times square root(x)

[spring steeple]

sqrt(x)

(1 is the multiplicative identity)

[pearl knoll]

ok

and then square root of x+h times sqrt(x) is

?

[spring steeple]

Yes

[pearl knoll]

what is it

[spring steeple]

Just keep it sqrt(x+h) times sqrt(x)

No need to simplify now

[pearl knoll]

oh ok so you dont multiply

ok now for 1 over sqrt(x)

[spring steeple]

sqrt(x) will become the denominator and 1 when multiplied by the numerator won't change it

[pearl knoll]

i got square root of x+h over square root of x times x+h

is that right?

[spring steeple]

Yes and there is minus something as well

-sqrt(x+h) over sqrt(x) sqrt(x+h)

[pearl knoll]

yes

[spring steeple]

Now take 1 over sqrt(x)sqrt(x+h) as common factor

[pearl knoll]

so we have sqrt(x) over sqrt(x+h) times (x) - sqrt(x+h) over square root of x times square root of x+h)

[spring steeple]

Yes take 1 over sqrt(x) times sqrt(x+h) as common factor now

[pearl knoll]

why 1?

[spring steeple]

Because you can't factor out the denominator alone

You need 1 over ...

Also you can do that because 1 is the multiplicative identity so it doesn't change the expression

[pearl knoll]

im confused

[spring steeple]

1 times a is a

[pearl knoll]

yes

but im confused as to where you are going

[spring steeple]

So you can factor out the 1

[pearl knoll]

with this

[spring steeple]

Ohh

We need to factor out this and then take the conjugate of the numerator (of the numerator of the original expression) but no need to worry about that now

Just keep going

[pearl knoll]

i think it would be easier if you showed me

like on paper or online

[spring steeple]

Ahh okay just a sec then

$\frac{\frac{\sqrt{x}}{\sqrt{x}\sqrt{x+h}} - \frac{\sqrt{x+h}}{\sqrt{x}\sqrt{x+h}}{h}$
$\frac{\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x} \sqrt{x+h}}}{h}$

[warm shale]

Normed
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[pearl knoll]

this is what i have so far:

[spring steeple]

Yes factor out 1 over sqrt(x) times sqrt(x+h) now

[pearl knoll]

for the first fraction>

?

[spring steeple]

Factor out the term 1/sqrt(x)sqrt(x+h)

[pearl knoll]

where are you getting the 1 from

[spring steeple]

1 times sqrt(x) = sqrt(x) and 1 times sqrt(x+h) = sqrt(x+h) so you can factor it out

[pearl knoll]

im literally so confused

we have a common denominator and thats sqrt(x+h) times sqrt of x

so now we just need to subtract sqrt of x-sqrt(x+h)

[spring steeple]

Yeah (implicit in this step is the factoring out step) but if you get that nvm then

[pearl knoll]

so what is sqrt of x-sqrt(x+h)

[spring steeple]

You can't simplify that directly

Multiply top and bottom by the conjugate of that

[pearl knoll]

whats the conjugate

[spring steeple]

sqrt(x) + sqrt(x+h)

[pearl knoll]

ok i get that

but why the conjugate

this is taking me forever

[spring steeple]

So that we can apply difference of squares

a^2 - b^2 = (a+b)(a-b)

And squaring these terms will get rid of the sqrt

[pearl knoll]

this is taking me forever

im not understanding anything

[spring steeple]

Ahh really sorry, but this is the only way I know of solving such problems

[pearl knoll]

it literally is taking us an hour to do this. there has to be an easier way

[spring steeple]

Let's see if someone else can help you better than me with this
<@&286206848099549185>

[pearl knoll]

ok

maybe i should ask in a different help channel

.close

[obtuse pebble]

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[sinful nova]

?

[spring steeple]

Ahh read the context I suppose

[spring steeple]

Close this channel (unless you have a question to ask), they are getting help in a another channel now

[sinful nova]

oh okay

.close

[obtuse pebble]

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[severe granite]

I tried to use a recurrence to solve this question but I am stuck on this step (last pic), idk if I'm even on the right path

<@&286206848099549185>

[polar fossil]

i feel like you've done it?

z • z^n = z^{n+1}

[brave bramble]

You should mention a property of * somewhere in your proof

You are implicitly using one, when you take (z^(n+1))* and turn it into (z^n)*z

[severe granite]

Oh I see

Thanks

.close

[obtuse pebble]

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[cold brook]

can someone help with this

[trail musk]

Use the ab sintheta formula

[cold brook]

how will that prove that it transforms as a vector?

what do I have to technichally do? Can you do the overview, i might be able to do it from there. i just dont understand the question

[thin star]

What does it mean for something to transform as a vector?

[obtuse pebble]

@cold brook Has your question been resolved?

[obtuse pebble]

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[cold brook]

the length and angle should be conserved?

[warm canopy]

What is "the angle" of a 3d vector?

[spring steeple]

transforms as in linear transformation ?

[cold brook]

AdotB/|a||b| ?

[warm canopy]

That's the angle between two vectors

[cold brook]

oh yea

can we find the angle

with each axis

[warm canopy]

the question is very unclear imo, A x B is a vector by definition, what does it mean for a vector transform as a vector?

[cold brook]

yea thats what is confusing me too

i have no idea

what they mean

[warm canopy]

my best guess is they mean, if you rotate A and B by some angle about some axis, the same thing happens to A x B

but thats really just my interpretation and what i think is a perhaps interesting question

[tardy epoch]

this maybe?

https://www.math.fsu.edu/~quine/MB_11/4_rotations.pdf

[warm canopy]

so yeah this

[cold brook]

i dont get it tbh

[tardy epoch]

Have you learned what rotation matrices are

[cold brook]

cos Q -sinQ
sinQ. cosQ

this thing?

[tardy epoch]

that's true for two dimensions

[tardy epoch]

are A and B 2d vectors?

cross products are defined for 3d vectors only

[cold brook]

we havent done rotation in 3d

[tardy epoch]

yea i'm not sure how you'd do this if you don't know anything about rotation matrices in 3d

[thin star]

@cold brook did you guys at least cover the index notation for the application of an matrix onto a vector?

[cold brook]

Wait can you show me an example of that? Tbh we haven't been taught shit. All this Levi civita and delta stuff, my prof just straight up wrote the rules and finished in 10 mins. And we just went over vector and co-ordinate system rotation in 2d and derived a rotation matrix for it

Also she uploaded this recently though didn't teach it. Do we have to use this? If then I will go through it

[tardy epoch]

can you just screenshot all the 3d matrices and upload

[cold brook]

Yess give me a moment

It's mostly for n dimensions

[obtuse pebble]

@cold brook Has your question been resolved?

[thin star]

That's all I meant. That's the expression for the i'th element of some matrix x after applying a rotation R

What I'm guess is that you need to do this for some vectors x and y. Then show that you get the same thing for the cross product

Basically the identiy Riemann sent

#help-10 message

[cold brook]

i remember we were taught this but a TA taught it and we didnt get shit. i will try to learn it from youtube

thankyou though

[fickle surge]

I’m just wondering if there an easier way to decide when to use the typical volume slicing method, washer disc method, or shell method just based on what i formation is given in the word problem.

[obtuse pebble]

@cold brook Has your question been resolved?

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[vapid laurel]

If a and b are two real numbers satisfying a+2b=50 what is the max value of a*b? solve using the AGM inequality

[vapid laurel]

I saw the solution but I don't understand why they used a and 2b for x and y in the inequality

2b specficaly

[polar fossil]

i mean probably because we have a + 2b = 50

so we have (a + 2b) / 2 = 25

[vapid laurel]

.close

[obtuse pebble]

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[steady patrol]

hi can you help me with this

i didn't understand the step

someone explained this to me but i forgot how

[thin star]

The simplified the other factor

Multiply and divide by 2

classic trick

[obtuse pebble]

@steady patrol Has your question been resolved?

[obtuse pebble]

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[hard imp]

so this is the graph for the problem. I dont know if i answer C correctly and im not sure how to D

[hard imp]

i also need help with 2 more questions after these if anyone is willing to help me

[obtuse pebble]

@hard imp Has your question been resolved?

[hard imp]

<@&286206848099549185>

[lone echo]

on C, i'd answer with the limits and the value of the function.
Your answer is "it's not continuous because it isnt."

[hard imp]

Okay, so should i put lim x->3 f(x) = -1 ?

[lone echo]

that's only part of what you need

[hard imp]

oh okay. Do you know how to do part D?

[lone echo]

remember that for a function to be continuous at a point, you need both lateral limits and the value of the function at the point to be defined and be equal

[hard imp]

<@&286206848099549185>

[hard imp]

i got help with C, im still trying to figure out D

[obtuse pebble]

@hard imp Has your question been resolved?

[hard imp]

<@&286206848099549185>

[hard imp]

.close

[obtuse pebble]

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[thin star]

That's some sick prison tattoo

[serene fossil]

fr fr

[thin star]

I'm not gonna lie, I have never seen this before

[serene fossil]

if you wanna apply this method for ranges outside of 0 to pi/2 you can always remember the cast rule,

[obtuse pebble]

@open forge Has your question been resolved?

[serene fossil]

Basically this rule says that the values of the trig functions are essentially the same and periodic about the circle just with different signs and the CAST mnemonic tells you which sign to use

In my opinion it’s easier to remember what the sine and cosine represent

The sine of an angle x is the ratio of the side length opposite x to the side-length of the hypotenuse of a right triangle

Remember the phrase SOH CAH TOA maybe

Then if you remember this you can draw the triangle in the correct quadrant and not have to worry about the CAST rule

You can just look at the circle and realise what needs to be positive and negative

[open forge]

alright, my instructor is trying to get us away from using the unit circle because it’s handy to know these quickly

[serene fossil]

That’s an interesting approach

I think anything you learn in trig should be about understanding and not little tricks of memorization since memorizing this stuff is worthless anyways

You never actually need to evaluate a trig function by hand

If there were a case where you need to, why would you not use a unit circle if it’s easier

And if time is a problem then calculators exist

But I guess it’s your instructirs choice

[open forge]

instructor is against referencing a unit circle and using calculators for trigonometry/logarithms...

🤷‍♂️

[thin star]

wait a second

what do you mean "against reference a unit circle"?

I can understand being against calculators

[open forge]

like looking at it directly

i.e. on paper or online

has to be in our head...

[thin star]

Even if you drew it yourself on paper?

[daring rock]

you really just need to memorize the first quadrant, you can figure everything else out on the fly by reflections

[thin star]

That's insane, but yeah go do flashcards or something

[open forge]

ya ik first quadrant pretty well

im just bad at mentually visualizing reflections

[daring rock]

I just think of it like, sqrt(3)/2 > sqrt(2)/2 > 1/2, so if you can picture it, you can figure out if you need big/med/small x value and big/med/small y value

for example 2pi/3 is near the top of the second quadrant

quadrant 2, small x value, big y value

(-1/2, sqrt(3)/2)

[open forge]

oh ill try to think of it like that

[obtuse pebble]

@open forge Has your question been resolved?

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[pale scaffold]

Question tells me to find the specific value for this

[pale scaffold]

No idea how to begin

[thin star]

Okay

Are you aware of the log and exponetial identities>?

[pale scaffold]

yup

[thin star]

so what is $e^{\log x}$?

[warm shale]

TooManyCooks

[thin star]

sorry, when I write log I mean natural log

[pale scaffold]

oh ok

[thin star]

the best kind of log

[pale scaffold]

so e^ln (x)

[thin star]

yes

what is that equal to

[pale scaffold]

well it would be ln (e^ ln x)

i think so

[thin star]

where did the extra ln come from?

You just want $e^{\ln x}$

[warm shale]

TooManyCooks

[pale scaffold]

ok

ln (e) + ln (x)

[thin star]

Not quite

[pale scaffold]

trying to do the product rule

[thin star]

Okay how about this

You know exponentiating with e and taking ln are inverse operations right?

Or no?

[pale scaffold]

no

[thin star]

Okay

Tell me what properties of ln you know

Let's start there

[pale scaffold]

well product property

quotient property

power property

and thats kind of all i know

[thin star]

Okay. There's also this property

$\text{blob} = e^{\ln \text{blob}}$

[warm shale]

TooManyCooks

[thin star]

you can replace blob with anything you want

[pale scaffold]

ok

[thin star]

the reason i'm telling you this is because you want to use that trick for your problem

[pale scaffold]

so if we did e^ ln (x) it would equal to x

[thin star]

exactly!

[pale scaffold]

ah ok

i guess i was overcomplicating it

[thin star]

Ok let's look at the exponents first

You have $e^{-\ln 2}$

[warm shale]

TooManyCooks

[thin star]

Focus on the -ln 2

[pale scaffold]

ok

[thin star]

use the power rule

[pale scaffold]

ln 2^-1

[thin star]

GIve me the expression for what $-\ln 2$ is

[warm shale]

TooManyCooks

[thin star]

using the power rule I mean

[pale scaffold]

ok

[thin star]

Let's pretend we don't know how to deal with exponents with - ln

[pale scaffold]

would it be ln (2)^-1

[thin star]

Perfect

[pale scaffold]

ok

[thin star]

So let's use the identity trick

x = e^ln x

[pale scaffold]

so it would be -2 = x

no

1/2

1/2 = x

[thin star]

yes exactly

good job

[pale scaffold]

thanks, its pretty easy actually

[thin star]

Good! I'm glad you think so

You can play more tricks like that

let's do the second one

[pale scaffold]

so now i use the same property for question 2

[thin star]

I'll let you handle this one first and see how far you can go

[pale scaffold]

ok

so i got e^ln(3)

because i did (ln e^3) and simplifyed that to 3

[thin star]

ok

then what

[pale scaffold]

so with that x =3

by just looking at our property

[thin star]

yep

exactly

1/2 and 3

[pale scaffold]

alr

ive actaully got a another question

[thin star]

sure thing

what is it?

[pale scaffold]

its this here

[thin star]

okay let's start with 60

[pale scaffold]

ok

[thin star]

what do you do

[pale scaffold]

im assuming i use the power property first

so it would be 2x (ln (1+ e) ) + 2x ( ln (1+ e) )

[thin star]

You can't do that here

[pale scaffold]

hmm ok

wait

[thin star]

You can't do it because there's a sum

[pale scaffold]

can we cancel it out

[pale scaffold]

wait whats the sum

oh wait

theres a - in between

[thin star]

No

[pale scaffold]

so we can use the quotient property?

[thin star]

I meant inside the log

[pale scaffold]

oh

product property inside the log?

[thin star]

But yes you do need the quotien property here

YOu can't do product property INSIDE the log

only outside

[pale scaffold]

ok

well what do i do then? I cant do the product property, nor the quotient property first, and the power property

[thin star]

Let's focus on what we CAN do

[pale scaffold]

wait is it possible to do the one to one exponent property

[thin star]

You have there the difference of two logs

Let's start with that

[pale scaffold]

ok

[thin star]

What's the rule for $\ln a - \ln b$?

[warm shale]

TooManyCooks

[pale scaffold]

you divide it

ln (a/b)

so we can perhaps replicate this here and do ln (1 + e^2x / 1 + e^-2x)

[thin star]

Right

So can you do anything to simplify that thing inside the log?

[pale scaffold]

we could

maybe using the power property?

no i meant the product property

[thin star]

No

ignore the log

just focus on that expression

[pale scaffold]

ok

[thin star]

can you simplify it?

Here's a hint: eliminate any negative powers

This is just algebra at this point. Ignore the logs

[pale scaffold]

yeah my brains gone blank

nvm

to get rid of the -2x we do 1/2x

[thin star]

$e^{-2x} = \frac{1}{e^{2x}}$

[warm shale]

TooManyCooks

[thin star]

use that

[pale scaffold]

alr

so it would be e^4x

after simplifying all of that

[thin star]

Uhh

What

Can you show me

What you ahve so far

[pale scaffold]

well i did 2x divided by 1/2x

i havent been writing down anything ._.

[thin star]

ok. you may wanna

could jbe scribbles

[pale scaffold]

ok

[thin star]

I need you to simplify the expression

$\ln\frac{1+e^{2x}}{1+e^{-2x}}$ is definitely something you can simplify

[warm shale]

TooManyCooks

[pale scaffold]

hmmm ok

[thin star]

Use my hint

[pale scaffold]

I think im a little too stupid to notice your hint ._.

[thin star]

do you see anyh negative powers in the expression?

plug my hint in there

[pale scaffold]

wait do 2x/2x cancel out and equal to one

oh wait ofc its gonna equal to one

[thin star]

There's no cancellation here

[pale scaffold]

ok

[thin star]

Look, if you're struggling doing this in your head, just write it down

[pale scaffold]

well im doing 2x multiplyed by 1/2x

[thin star]

$\ln\frac{1+e^{2x}}{1+\frac{1}{e^{2x}}}$

[warm shale]

TooManyCooks

[thin star]

There I did it for you

I want you to simplify that

[pale scaffold]

oh

[thin star]

that fraction in the denominator looks ugly

get rid of it

[pale scaffold]

to get rid of it we do 2x multiplyed by 1/2x?????

[thin star]

what?

I didn't multiply anything

I just rewrote the negative exponential into a fraction

[pale scaffold]

yeah

but im trying to get rid of the fraction

[thin star]

Ah okay. You play with it for a bit

[pale scaffold]

i dont play right now with fractions 😭

[thin star]

Okay. Here's a hint

$\left(\frac{e^{2x}}{e^{2x}} * 1 + \frac{1}{e^{2x}}}\right)$

oops

[pale scaffold]

i feel... stupid

[thin star]

goddamnit

lmao gimme a sec

[pale scaffold]

ok

[warm shale]

TooManyCooks
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[thin star]

I just did the trick where you multiply by 1 in a fancy way

[pale scaffold]

ok

so here it would be (e^2x divided by e^4x) + 1

[thin star]

Can you type or write it up?

[pale scaffold]

ok

ill write

[thin star]

or you can type it in an online website, then take a screen shot

either way I just want to know what you have so far

[bold hound]

in the pallalelegram illustrated here, determine the coordinates of point P relative to the reference points

^this is the translation

[thin star]

do it in a new channel lmao

[bold hound]

bruh

https://tenor.com/view/johnny-crying-secteur-nord-berlingot-gif-13277851

[pale scaffold]

this is what i did

[thin star]

I want you to multiply just 1

not the whole thing

[pale scaffold]

oh

well then it would be (1 + 1/e^2x) no change to it since multiplying by one

[thin star]

What I was getting at is the fact that you can group then in a way that solves your problem

Don't forget $\ln\frac{1+e^{2x}}{1+\frac{1}{e^{2x}}}$

[warm shale]

TooManyCooks

[pale scaffold]

hmmm ok

[thin star]

You're just rewriting the denominator

[pale scaffold]

yeah

so now with this denominator rewritien we can simplify it

[thin star]

It simplifies a lot

so what do you get

[pale scaffold]

since we are dividing we have to change the fraction into its inverse

inverse isnt the right word for it

[thin star]

$\ln \frac{1+e^{2x}}{\frac{1+e^{2x}}{e^{2x}}}$

[warm shale]

TooManyCooks

[thin star]

does it help if I write it that way?

[pale scaffold]

yeah

but how would we get that for the denominator

[thin star]

What do you mean??

[pale scaffold]

wait did you rewrite the denominator

[thin star]

This is all just rewriting

I haven't simplified yet

[pale scaffold]

oh ok

i understand what you did to rewrite it

[thin star]

Okay

[pale scaffold]

you divided by e^2x

[thin star]

I need you to be 100% sure you do

[pale scaffold]

is this it?

[thin star]

Divided what

[pale scaffold]

e^2x

the deonimator

deonimator

denominator

denominator divided by 2x

[thin star]

I did not divide by 2x

There are only constants and exponentials here

[pale scaffold]

ok

Well what did you do then? If you didnt divide by 2x

[thin star]

I used the converted the 1 into $\frac{e^{2x}}{e^{2x}}$

[warm shale]

TooManyCooks

[thin star]

and grouped it with the $\frac{1}{e^{2x}}$

[warm shale]

TooManyCooks

[pale scaffold]

ok

[pale scaffold]

wait can you explain this step

[thin star]

$1 + \frac{1}{e^{2x}} = \left(\frac{e^{2x}}{e^{2x}}\right) + \frac{1}{e^{2x}} = \frac{e^{2x} + 1}{e^{2x}}$

[warm shale]

TooManyCooks

[pale scaffold]

ok

i understand it

so what do we do next after this step

[thin star]

Simplify the fraction inside the ln

[pale scaffold]

this is what i have

,rotate

[pale scaffold]

so now we simplify it

the top part of it will cancel out

getting us 1/e^2x

[pale scaffold]

forgot to put an x at the e^2

[pale scaffold]

is this right?

[thin star]

missing an x

[pale scaffold]

yeah i know

[thin star]

But you just wrote what I wrote

[pale scaffold]

i now have ln (1/e^2x)

[thin star]

Good!

wait

no not quite

[pale scaffold]

ok

[thin star]

almost though

you have a fraction of a fraction.

[pale scaffold]

yeah

and i canceled out the top part of the fraction

since its just 1 + e^2x / e^2x + 1

its the same thing

[thin star]

$\frac{1}{\frac{1}{e^{2x}}}$

[warm shale]

TooManyCooks

[thin star]

That's what you have

simplify that

[pale scaffold]

oh ok

thats gonna be 1/2x?

the top part cancels out?

or is their a another way of doing this

so yeah it will be e^1/2x

i now have ln (e^1/2x)

so now we use power property

-ln(e^2x)

[pale scaffold]

@thin star is this right?

[thin star]

Where did you get the minus sign

[pale scaffold]

nvm

it would be -2x(ln e)

[thin star]

there should not be a minus sign at all

[pale scaffold]

ok

[thin star]

$\frac{1}{\frac{1}{e^{2x}}} = e^{2x}$

[warm shale]

TooManyCooks

[pale scaffold]

just relized when i checked my answer sheet that it says only 1+e^2x / 1+e^-2x

so it doesnt require you to simplify to this extreme amount

[pale scaffold]

this here is the solution in the answer sheet

[thin star]

that's kinda dumb considering this is simplifiable to 2x lmao

[pale scaffold]

yeah

so now we have that ln (e^2x)

do power property

and we get the answer 2x

well that was 60 now its time for 61

no idea how to do this one

[thin star]

ok

do you know the change of basis rule?

[pale scaffold]

yeah that rule

[thin star]

$\log_{a} b = \frac{\log_c b}{\log_c a}$

[warm shale]

TooManyCooks

[thin star]

this one

[pale scaffold]

so we use that rule for log to the base p of e

yup

so we use this rule for both the first natural log and second log

[thin star]

yes

let c = e for both

doesn't really matter i guess

[pale scaffold]

yeah

[thin star]

just my personal choice

[pale scaffold]

ok

im sending a picture

,rotate

[pale scaffold]

so if we simplify both the natural log and the log, it will equal to 1

[thin star]

The first two factors should cancel, yes

[pale scaffold]

yeah so the first two equal to 1

[thin star]

so what do you do with the sqrt factor

the last one

[pale scaffold]

well

do you just do the exponential and logarthemic rule

[thin star]

I'll let you do it for now

should be a good exercise

[pale scaffold]

ok

its going to be sqrt of p

yup looks similar to the answer sheet

ive got a another question, its about solving for x

[thin star]

show me

[pale scaffold]

need some help on both

can we use the logarthmetic and exponential rule for 77

i beleive we can

[thin star]

yes

for 77 take ln of both sides

you can use the identity trick

[pale scaffold]

ok

i didnt do ln for both sides so i will do that now

what would be on the right side

i have got the left side of it but not the right side

I have gotten ln (6) on the right side

oh im dumb

i dont simplify it

7-4x = ln (6)

now it is -4x = ln (6) - 7

alr my answer is similar to the answer sheet

for 78 do i do the quotient property?

[pale scaffold]

@thin star would that be the correct approach?

[thin star]

no

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no

[thin star]

exponentiate it

[pale scaffold]

it is not

ok

alright i got the answer and its similar to the answer sheet

im going to keep this channel open while i do my homework, ill ping you if I need help

[thin star]

Yeah

sounds good

i'm helping like 2 other people right now haha

[pale scaffold]

noice

@thin star i need some help on two problems

I need help on 80 and 81

ngl dark mode looks weird after having light mode on for 1 year

<@&286206848099549185> Need some help here on 81 and 80

[thin star]

let e^x = y

[pale scaffold]

ok

e = 2

x = x-5

y=3

@thin star im kind of stuck on 81

[thin star]

Log sum property

then exponentiate

[pale scaffold]

can you show me log sum property

on texit

[thin star]

$\log a + \log b = \log ab$

[warm shale]

TooManyCooks

[pale scaffold]

oh the product property

so log 2^(x-5) = 3

do i log both sides?

i think so

i did it wrong

its 2^ log (x-5) = log (3)

[pale scaffold]

@thin star is that correct?

[thin star]

o shit i thought you were doing 82

sry

[pale scaffold]

oh

its fine

[thin star]

for 81 you just take log

both sides

then use power rule

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log (2^x-5) = log(3)

[thin star]

(x-5)log2 = log3

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x-5(log of 2) = log(3)

alr

now whats next

i got the correct answer

81 is done

@thin star now how do i do 80?

[thin star]

I thought you already did that

y = e^x

[pale scaffold]

you were talkign about exponentiating it

ok

[thin star]

No, I said y= e^x

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hmmm but how would we use that

can you show me the first step of using this?

[thin star]

Give me what expression you get in terms of y

[pale scaffold]

not quite sure what you mean by that

[thin star]

I asked if you could let y = e^x

You plug that into 80

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so we are substituiting?

[thin star]

Yes

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ok

how would you do it for e^2x

[thin star]

What is e^2x in terms of e^x

[pale scaffold]

well its just 2 times that

y^2

is it y^2

[thin star]

yes

[pale scaffold]

time to factor

[thin star]

Yep 😅

[pale scaffold]

y= 1, 2