help-23

[clever karma]

okay yayyy

how wuld u do 1/3lnx+2ln(3x-5)

ik u have to use power propetty

[empty gyro]

What's it equal to?

[clever karma]

nothing we just have to express as a single log

sorry forgot to mention that

[empty gyro]

then yes, use the power property

[clever karma]

so it becoms lnx^1/3+ln(3x-5)^2

and then i can use the like add propety or wtv its caled

so like lnx^1/3(3x-5)^2

can i simplify that any more

[empty gyro]

what you wrote is satisfactory

[clever karma]

wait is ln a log

okay so for the problem 2lnx-1/2ln(x^2-1)+3ln(x^2+1) i simplified it so ln x^2/(x^2-1)^1/2 + ln (x^2+1)^3 is that right so far???

and then it bcomes the first part times x^2+1)^3

[heavy glade]

id help if i could understand anything, just write and send photos

[clever karma]

tysm! im just still a little stuck on understanding how to do ln(ln x)=2

[heavy glade]

that would mean e^2 = lnx which means e^e^2 = x

[clever karma]

wait it becomes like that because u make e on both sides

but one of the lns cancel out the e

so it just becomes e^2=lnx

but how does it become e^e^2

like do u just divide by ln?

[heavy glade]

say lna = b

[empty gyro]

Yes. ln is the natural log

[heavy glade]

bro i got hw, plz help them

[marsh gorge]

hi

[clever karma]

🥲 🥲 🥲

[empty gyro]

yes

[clever karma]

tysm for helping me understand those problems i actually get them now

im just confused on the ln(ln x)=2

i get how to do lnx=e^2

[empty gyro]

First, take e exponent of both sides: If $\ln(\ln x)=2$, then $e^{\ln(\ln x)}=e^2$

[flat frigate]

SWR

[clever karma]

and then e and ln cancel out

bc they inversw

[empty gyro]

leaving you with what?

[clever karma]

okay so now (ln x)=e^2

[empty gyro]

Take e exponent of both sides again

[clever karma]

omg

OMG

so its now e^lnx=e^e^2

[heavy glade]

[clever karma]

leaving u w x=e^e^2

OMG

savior

so anytime theres an ln just put it to the cpower of eto cancel out

[empty gyro]

Just be sure you write $e^{e^2}$ and not $(e^e)^2$. They mean different things.

[flat frigate]

SWR

[clever karma]

omg tysm youve genuinely taught me more than my teacher has all semester

[empty gyro]

good luck on the rest of your semester

[clever karma]

TYSMMMM!!!!

.close

[safe radish]

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[clever karma]

.reopen

[safe radish]

✅ Original question: #help-23 message

[clever karma]

heh...not questionless for long

how do i do e^x=4

ik its supposed to be simple but uh

[empty gyro]

just ln on both side

[raven hearth]

x= ln(4)

[clever karma]

oh my im such a chud

thanks guys

.close

[safe radish]

Channel closed

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Use .reopen if this was a mistake.

[clever karma]

.reopen

[safe radish]

✅ Original question: #help-23 message

[clever karma]

HELLO

how do i do 2^x-5=3

im confused now that it doesnt have an ln or e

do i just do 2(x-5)=3

[empty gyro]

Take log of both sides

[clever karma]

oh fart

[warped reef]

2^x=8

[clever karma]

how did u get that

[warped reef]

X=log8, base 2

[warped reef]

Add 5 on both sides

[empty gyro]

$2^{x-5}=3$ or $2^x-5=3$?

[flat frigate]

SWR

[empty gyro]

!nosols

[safe radish]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[clever karma]

damn so just log base 2^x=log base 8

[clever karma]

first

[empty gyro]

👆

[warped reef]

Ohh, 8 misinterpreted the q

[warped reef]

I*

[clever karma]

btw im finding the solution of each equation to four decimal places

so just log base 2^x=log base 8?

but like does the x come down i dontget it

[warped reef]

Not at all, if the 1st one is ryt

2^(x-5)=3 is the q?

[clever karma]

yes

[empty gyro]

@clever karma this is why parentheses are important

[clever karma]

i need to find the solution so it should be like a decimal point

SORRYY GUYS

[warped reef]

Then wt i said is wrong haha

[empty gyro]

Try writing 2^(x-5)=3 next time

[clever karma]

okay fsfs

[warped reef]

Then take log on both sides(base2)

So that 2^ is removed

[clever karma]

wait so can i just make the right side log base 2 of 3?

i can make the base wtv i watnt right

[warped reef]

Yes

[empty gyro]

Yes. Same as you did with the natural logs in earlier problems

[clever karma]

so when i have a problem like that if i want i can make it a log

even when it has no ln and e

[warped reef]

Ya, but the goal here is to eliminate 2^

[clever karma]

okay nice

so now its x-5=log base 2 of 3?

[warped reef]

For simplicity. U can use ln too

[warped reef]

Yess

[clever karma]

so its just log base 2 of 3+5

[warped reef]

Parentheses, but ya

[clever karma]

tysmsmsmsms

.close

[safe radish]

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[clever karma]

.reopen

[safe radish]

✅ Original question: #help-23 message

[clever karma]

help

e^x+4e^(-x)=5

i got as far as e^x+e^(-4x)=5

idk of thats right

do i combine the e powers?

[heavy glade]

can you?

[clever karma]

not sure ngl

i dont think so if ur not mulriplfying right

[heavy glade]

yep

uk what negative power means right

[clever karma]

so i becomes ln^x+ln^(-4x)=5

fraction right

[heavy glade]

like the reciprocal of the thing without the negative

so a^(-b) = 1/a^b

[clever karma]

i just dont know how to turn this into deicmals

decimals

[heavy glade]

idk how you got these mate

[clever karma]

oh fart

did i do smth wrong

i fdont get how to simplfiyit

[heavy glade]

idk i just dont get it

[clever karma]

me too

[heavy glade]

id first turn the negative power into a "fraction"

basically its reciprocal form

then see what i can do

[clever karma]

okay so e^-x---> 1/e^x

[heavy glade]

yep

[clever karma]

where do i go from e^x+4(1/ex)=5

[heavy glade]

uk how to solve quadratics right?

[clever karma]

si

[heavy glade]

alright

[clever karma]

omg no dont tell me this is a QUADRATIC

[heavy glade]

here take LCM

[clever karma]

e^x

[heavy glade]

so turn them into like fractions

[clever karma]

factor it out?

oh

wait im so sorry wdym vy that

[heavy glade]

you have e^x and 4/e^x

[clever karma]

si

[heavy glade]

make these two into like fractions

uk what like fractions are right

[clever karma]

uhm

okay wiait yes

how do i do that again

IM SORRY I FORGET THINGS

[heavy glade]

so multiply the e^x by e^x/e^x

then both of the terms will have denominator e^x

[clever karma]

oay so 1/e^x+4(1/e^x)=5

[heavy glade]

did you learn in spanish? if so i could explain terms like denominators and all that

[clever karma]

no

[heavy glade]

no (e^x)^2 / e^x + 4/e^x = 5

[clever karma]

im soryr i just say si

[heavy glade]

alright

[clever karma]

okayokay

what do i do from here

[heavy glade]

quadratics 💀

[clever karma]

OH

SHIIIIIHEHHAHHHAHA

LMFAO

LMFAO

okay i see sorry

im a bum

tysm

[heavy glade]

yep

[normal moss]

That 💡 moment

[clever karma]

.close

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• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
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[mortal sandal]

Try multiplying it by itself once by hand, the inner and outer blocks essentially act separately

in fact this can be Jordan-ized with the right permutation matrix

[dim socket]

I have not read it yet

should I have to trap the pattern?

[mortal sandal]

I mean it's basically two separate matrices

the first and fourth rows/columns and the second/third

But the easiest way is probably just to compute a few powers by hand and then see there's a simple pattern

[honest perch]

do you know about diagonalization

[safe radish]

@dim socket Has your question been resolved?

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• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[potent oar]

"define where f(x) exists"
hm... okay i get the bottom part, that's x+2>0 so x>-2, that's fine, but the top part...
the argument of log must be greater than 0, so e^-4x - sqrt(e) > 0, so far so good, but then
e^-4x > sqrt(e)... yeah i'm lost now... how do i proceed?

[potent oar]

i mean, i can square both...
e^-8x > e... that doesnt get me much further...

[analog quartz]

try taking log?

[potent oar]

1^-8x > 1?

🤷‍♂️

[analog quartz]

wdym?

[potent oar]

is that wrong?

[versed wave]

yes

[analog quartz]

i meant you can take log on both sides

[potent oar]

if i rememer correctly log(e^x) is... wait what was it...

oh okay wait

so... -8x > 1? idk...

[analog quartz]

yeah

[potent oar]

oh okay yeah the solution...

x<-1/8... so... C

[analog quartz]

you can also say since e^-8x > e, then than implies -8x > 1, since the base is same here

[potent oar]

and since i multiply x by a negative number i have to invert the inequality, right?

[analog quartz]

yeah

[potent oar]

okay perfect, thanks

.close

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[lofty kelp]

??

[fading mural]

hint: AM-HM inequality

[lofty kelp]

oh , how ??

we arnt suppose to maximize anything here right ?

thats what we use AM - GM inequality for , generally

also i am not getting that summation

[fading mural]

yes, but you will surprised by how it is used here, tho I want you to figure it our your own

[fading mural]

bounding value

[lofty kelp]

ohh

[fading mural]

-# 🥀 the main reason is that I'm playing mc

[lofty kelp]

lol

can u give a little hintt

from where to start

ill solve it

[fading mural]

, start by finding the interval where both roots of f'(x) possibly be

ig

compare them to alpha, beta and gamma

[lofty kelp]

ohk let me try

[analog quartz]

i think ive gotten the answer but im not sure

but yes am gm is used here in the end

ill try to give a hint

[lofty kelp]

sure

[analog quartz]

||using the information given, you can find a value of beta in terms of the coefficients, and same for alpha + gamma. once you sub that in, apply am gm inequality||

[analog quartz]

i think its intuitive here since you have the GIF

[queen ingot]

i found || b^2 = ag ||, is that correct?

[lofty kelp]

oh , yea

[queen ingot]

oh i forgot something

my leading term coefficient

[analog quartz]

no idts

[queen ingot]

oh you want the bound for || the greatest integer function, don't you? ||

[analog quartz]

yeah

[lofty kelp]

i got it

thanks

[analog quartz]

you can bound the expression inside

[lofty kelp]

6

[analog quartz]

whats teh answer?

[lofty kelp]

.

gif is k is 0

wo , 2^0 , 6 times

in summation

[analog quartz]

yeah

[lofty kelp]

.close

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[proven smelt]

for a graph like 1/x^2, is the range >0 or R{0}?

[jade charm]

it is greater than 0.

[safe radish]

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[thin stream]

hippity hoppoty this chanel is my property
are those two 75 degree angles correspodning?

[thin bridge]

the one with the red arrow, no

[thin stream]

the heccin

how are the two diagrams diffierent?

[thin bridge]

these are two separate lines

[marsh gorge]

yes if that vertical line is parallel to the north

and'

[thin stream]

trhat line is js supposed to be an extension of AB

duuuuuuuuuuuuuuuuude

[marsh gorge]

the other line is continous to ab

[marsh gorge]

then yes

[thin bridge]

if its intended as a direct extension it should appear as such

[thin stream]

my drawings are poopers

[thin bridge]

if it is, then yes, those angles would be corresponding

[thin stream]

sorry

oh

thank youi

holy shiz

imma keep this open

im gonna actually do the real question and get it checked by one of yallz

since i dont have access to the answer key

could some kind hearted young soul checkt his for me

i hope my handwriting isnt too shitty

[thin bridge]

c is incorrect

note that the bearing is the angle measured clockwise from north

[thin stream]

oh right

clockwise

oops

[thin bridge]

in part ii), that dotted line isn't facing north/vertical

[thin stream]

im going a little crazy

ive been doing maths all day

can i get a clue for cii?

@thin bridge

[thin bridge]

firstly that won't be a right angle up top

[thin stream]

waaaaaaaaaaaaaaat

it looks pretty durn

perpendicular to me

theta is the bearing im trying to find btw

[thin bridge]

what you drew earlier (which doesn't really help) would be representative of something perpendicular

extend that vertical line further

[thin stream]

the dotted line?

[thin bridge]

yes

[thin stream]

i did

[thin bridge]

that'll be another vertical/north line
and you can apply corresponding angles property again to get another 75° angle

[thin stream]

let me see

[thin stream]

i am not cooking

i should revisit this stuff

again before doing bearings

it cant be helped
le test soon

[thin bridge]

note that you already know <ABC

[thin stream]

yeeeeeeeeeeeeeeeeeeees

[thin bridge]

getting theta from there is pretty much an application of exterior angle theorem
(angle sum of a line with angle sum of a triangle)

[thin stream]

that is not accurate of what the real digram looks like tho
i cant just join the north line of and the north line of A to make zat triangle
lik... do u get wht im trynna say

oh wait

thats still AC but just super warped looking

argh

[thin bridge]

you should make a cleaner diagram

if you're editing an image electronically
you should have access to line tools

[thin stream]

cleaner diagram

i gtg

do some stuffill

be back

theta smells like 75

b ut im not sure

no its not

br

brb

like 30 min

[thin bridge]

that'll be another vertical/north line
and you can apply corresponding angles property again to get another 75° angle
note that you already know <ABC

getting theta from there is pretty much an application of exterior angle theorem
(angle sum of a line with angle sum of a triangle)

[safe radish]

@thin stream Has your question been resolved?

[safe radish]

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[thin stream]

OHHHHHHHHH
75=67.8+theta

[thin stream]

oops

.close

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[thin stream]

ty lad

[safe radish]

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[clever karma]

.reopen

oh

[clever karma]

chu

i need more help w ln and logs

(ln x)^2-4=0

[manic patrol]

hi OP, what have you tried so far?

[clever karma]

adding 4 to the riht

so (ln x)^2=4?

and then square root

[manic patrol]

that would be correct.

[clever karma]

lnx=2

and then e of both sides?

[delicate shore]

±2

[clever karma]

wouldnt it be e^2

oh chud

[manic patrol]

that is but one answer.

[clever karma]

im supposed to solve for x like

in decimals

[manic patrol]

always remember that when taking the square root of both sides, barring any limitations of domains or ranges, both positive and negative solutions should be considered.

[clever karma]

oh my gohs

okaywait

so its e^+_2??

[manic patrol]

yes. but resolve them separately; otherwise this looks rather ugly-looking. and since you need them in decimal form anyway, you should consider them separately.

[clever karma]

okay cool cool tysmmm

and i have one more question

e^x+4e^-x=5

i cant figue it out

[main mural]

multiply both sides by e^x

[clever karma]

wait so that would become (e^x)(e^x+4e^-x)=5e^x

[main mural]

yes

[clever karma]

do i distribute?

[main mural]

yup

what do you get?

[clever karma]

wait so e^2^2x

[manic patrol]

after doing so, you may be able to notice something amongst the remaining terms.
if it isn't clear, you could try rewriting one of the terms using the power law of exponents to more clearly expose the pattern.

[clever karma]

oh shoot

uhmmmmmm

[main mural]

what is e^x * e^x

[clever karma]

e^xe^x?

okay uhmm

e^2x

[main mural]

yup!

and what about e^x * e^(-x)

[clever karma]

so e^2x+4e^-x=5e^x

or

waiy

[main mural]

you haven't multiplied 4e^(-x) by e^x

[clever karma]

u add them

so e^0

so 1

[main mural]

yess

[clever karma]

so that jus equals four?

[main mural]

indeed

[clever karma]

so e^2x+4=5e^x

[main mural]

yes

[clever karma]

do i sub 4 from both sides

[main mural]

"sub 4"?

[clever karma]

subtract

sorry

[manic patrol]

might be unproductive.

[main mural]

oh i thought you wrote substitute lol

[clever karma]

do i just take ln of both sides ???

[main mural]

no

move everything to the left side

i.e. subtract 5e^x from both sides

[clever karma]

dont tell me i have to factor

[main mural]

well sorta

[manic patrol]

you will have to factor in a way, yes.

[clever karma]

wait im sorry idk how to factor it

e^2x-5ex+4=0

[manic patrol]

this is where my earlier message comes into play.

try rewriting e^2x.

[clever karma]

oh wait so 2xe

omg

[main mural]

now note something

e^2x = ||(e^x)^2||

[clever karma]

wait maybe is it 2e^x

[main mural]

stop guessing 🥀

xd

[clever karma]

guys im sorry that was my educated guess

[manic patrol]

please consider the power rule of exponents!

[clever karma]

dent that mean that the power just comes down

[main mural]

yes but you can also bring it back up

[manic patrol]

$a^{bc} = (a^b)^c$

[clever karma]

so (e^x)^2 ---> 2(e^x)

[flat frigate]

trilunar arithmetic (Columbina)

[manic patrol]

this only applies if you have a logarithm attached!

[clever karma]

omg

wait

BRING THE OTHER ONE UP

[manic patrol]

$e^{x^2} \neq 2e^x$!!

[flat frigate]

trilunar arithmetic (Columbina)

[main mural]

no, you probably mean (e^x)^2 ---> e^(2x)

[clever karma]

oh yes i do

[manic patrol]

(for all x except x = 1.)

[clever karma]

interesting

wait so do i bring -5ex up?

or e^(2x) down

[main mural]

great and the thing is you can do it the other way
you have e^(2x) but that equals (e^x)^2

[manic patrol]

you will need to target the e^2x.

[main mural]

that's it

[clever karma]

okay so i bring (e^x)^2 down

[main mural]

no you keep (e^x)^2

[clever karma]

so e^2x

[main mural]

no

[clever karma]

im so sorry i gtenuienlt cant get it

[main mural]

what you had is
e^(2x) - 5e^x + 4 = 0
now what you want to have is
(e^x)^2 - 5e^x + 4 = 0

[clever karma]

okay i hav the second one

[main mural]

great!!

the cool thing about it is that it's completely in terms of e^x

so now you can let t=e^x

[clever karma]

oh so i have to replace t with e^x

[main mural]

and you get t^2 - 5t + 4 = 0

[clever karma]

and factor it like a quadratic

[main mural]

the other way around but yeah!

just solve t^2 - 5t + 4 = 0

[clever karma]

oh shoot okay WAI TYSMSMMSMSMS

i really appreciate

[main mural]

[clever karma]

so i just tyr to get a common factor of e^x

so i can change it into t like this

[manic patrol]

(also sorry to both artemetra and ella for intruding when arte has already gotten this under control. I'll step back.)

[main mural]

ur good

[clever karma]

thank you as well!!!

.close

[safe radish]

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[clever karma]

.reopen

[safe radish]

✅ Original question: #help-23 message

[clever karma]

oka wait

after i factor it

it becomes (x-4)(x-1)

so i get 4, and 1

[main mural]

yes

[clever karma]

then do i go back and put those in the og equation?

[main mural]

so that means t=4 or t=1

[main mural]

ehh nah

[main mural]

now you can put in e^x back

so you have e^x = 4 or e^x = 1

[clever karma]

OH so i have two equtions and i need t slve for both

[main mural]

yesss

exactly

[clever karma]

so (e^4)^2-5e^4+4=0

and (e)^2-5e^2+4=0

[manic patrol]

overcomplicating!

[clever karma]

oh fart

[manic patrol]

and also wrong.

[clever karma]

oh FART

[main mural]

what do you get when you solve e^x = 4

[manic patrol]

since you know that e^x = 4, don't you have a tool that can help you find an unknown exponent given a base and value? you know, one that starts with an L?

[clever karma]

log?

[main mural]

yess

[clever karma]

so log base e of 4?

wait im so soryr im dumb i dont evenknow

is it ln

[manic patrol]

yes!

[clever karma]

so lne^4^2

[main mural]

it's just x=ln(4)

you can also write it as 2 ln(2) but ln(4) suffices

so that's one of your solutions for x

[clever karma]

so ln(4)-5e^4+4=0?

wait so i plug it back into the WHOLE equation

[main mural]

why are you plugging back in in the first place?

[clever karma]

oh chud

[manic patrol]

why are we plugging it back in though?

[clever karma]

i dont know

do i just make it ln(1) and ln(4)

[main mural]

[main mural]

yes!

[manic patrol]

our job is to find x; unless you want to cross-check for some reason, you don't have to substitute any of your answers back in.

[main mural]

what is ln(1)? it's a special value

[clever karma]

so in these types of problems i get e^x, make them t, and find x and then make those x's lns

[clever karma]

0?

[main mural]

yes!

so x=ln(4) or x=0

those are your 2 solutions

you are done

[clever karma]

niceeee tysmmm

[main mural]

pretty much

[clever karma]

for e^x-6e^-x=1 is it the same thing

[main mural]

yes it's identical

get everything in terms of e^x, and then do t=e^x and solve the quadratic, etc.

that's the general algorithm

[clever karma]

so (e^x)^2-e^x-6=0?

so t^2-t-6=0

[main mural]

now here you'll actually encounter a little trap so be careful. what values of t do you get here?

[clever karma]

3 and -2?

[main mural]

yes

[clever karma]

so ln(3)

[main mural]

yes

[clever karma]

but u cant do ln(-2)?

so js ln(3)

[main mural]

indeed!

niceeee you got it

[clever karma]

tysmmsmsmsm

[main mural]

[clever karma]

really appreciate it yall!

.close

[safe radish]

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[clever karma]

.reopen

[safe radish]

✅ Original question: #help-23 message

[clever karma]

uhm...guys....

(ln x)^2+(ln x)-2=0

[main mural]

you can do same as before except let t=ln(x) this time

[clever karma]

oh tyssmsms

[manic patrol]

if you see something that looks like a quadratic it most likely is a quadratic and you can most likely treat it that way.

[clever karma]

wait so it becomes t^2+t-2

which is (t+2)(t-1) so js ln(-2), ln(1)?

[main mural]

all correct up until the very last point

remember what t is

t=ln(x)

so t=-2 means ln(x) = -2

[clever karma]

ohhhhhhhhh

it changes hee

[main mural]

so what is x?

[clever karma]

so x=e or e^2?

[main mural]

e or e^(-2)

but yes

[clever karma]

i mean -2

yesssss

okay tysmM!M!M!

[main mural]

[manic patrol]

[clever karma]

.close

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[fallen stream]

question need to be solved using laplase trasform
actually i stuck after taking laplase x'(0) is not given

[zealous ingot]

thats ok

you'll get x(t) as some function where one of the terms has x'(0) as a coefficient

at that point you can sub the value for x(pi/2) and find x'(0)

[fallen stream]

i didn't get Your point
are you saying to treat x'(0) as a constant till laplace inverse

[zealous ingot]

yup

[fallen stream]

thanks if any issue comes i will be back
.close

[plucky elk]

.close

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[carmine tulip]

5 \times 360^\circ = 1800^\circ \
\text{or} \
5^\circ \times 360^\circ = 1800^\circ

[carmine tulip]

$5 \times 360^\circ = 1800^\circ \
\text{or} \
5^\circ \times 360^\circ = 1800^\circ$

[flat frigate]

fort craft 🇵🇸

[carmine tulip]

Which one is it?

[lost jewel]

first one

[carmine tulip]

thanks.

[wet notch]

Yep

[carmine tulip]

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[dark tartan]

i have to say if and why $\left( \begin{array}{rrr}8&4&9\-3&1&-3\-6&-4&7\\end{array}\right)$ can be diagonlized or not. so my approach was at first glance set $|A-\lambda E|=$ but i did it like three times and everytime i got a different characteristic polynom. prob im just bad in algebra but the numbers there get really really big. the idea is to compare the number of eigenvectors needed for a certain eigenvalue and if it is equal to the number of how often that specific eigenvalue is an eigenvalue. if you know what i mean.

[flat frigate]

Quark

[plucky elk]

!show

[safe radish]

Show your work, and if possible, explain where you are stuck.

[dark tartan]

My first and second try. With -lambda³ +16lambda² i'm quite confident. But the rest is problematic

*forgot. The determinant ofc has to be set equal to 0.

it is like figuring out when i can say my determinant is correct. guessing with such big polynomials is absurd

btw that was an exam exercise from 2016 without aids. so it should be solvable in like max. 12 min without a calc

[mortal sandal]

You did 3*64 = 202 near the end of first page

If you fix that your answer will be right

[dark tartan]

interesting. i should get -x³+16²-125x+278 =0 which is not quite better. like if i just try to get the zeropoints with a CAS i still get weird looking values. i dont really beliefe that is the way :(

ok i didnt want to but i looked it up. it is correct

i guess thx though im confused with that exercise. i can really not get those eigenvalues by guessing

they are even complex :')

[plucky elk]

you shouldn't be guessing. just fix the error dreyuk told you

[dark tartan]

i did. and it is correct.

[plucky elk]

so why are you using CAS

[dark tartan]

cause i have to tell if A can be diagonalised. and for that the number of eigenvectors for a eigenvalue has to be equal to the number of times this eigenvalue exsists. but my idea is that at this point it doesnt really matter what the eigenvalues even are. if i know that there are three, then i know that every eigenvalue has exact one eigenvector and A can be diagonalised

i try it another time. thx anyway

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[mortal sandal]

well the derivative is always negative so there's only 1 real root

so it's only diagonalizable over C

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[bleak notch]

Is this correct (question 3) because im trying to verify my answer with chatgpt but its saying the direction should be 33.6 instead of 26 or is there multiple ways to do this

[solar hazel]

good handwriting

[lost jewel]

,rccw

[solar hazel]

wait maybe i already told you that idr

[bleak notch]

Thanks🥹

[safe radish]

@bleak notch Has your question been resolved?

[tardy mango]

This is why you don't trust AI. You're fine.

[bleak notch]

ok tysm

.close

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[tardy mango]

[safe radish]

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[clever karma]

.reopen

.reopen'

.reopen

.open

oh fart

[clever karma]

2^x+3^x=-1

[plucky elk]

can x be a complex number

[clever karma]

no im doing like logs and e and ln

[plucky elk]

then i recommend taking a picture of the original question and sending it

[clever karma]

okay! one second!

g

[plucky elk]

,rotate

[plucky elk]

do you know that 2^x and 3^x are both always positive?

,w plot 2^x

[flat frigate]

Failed to get a response from Wolfram Alpha.
If the problem persists, please contact support.

Feb 19, 2026, 9:07 PM

[clever karma]

like do i know in thecontext of the problem orrr is that like factual?

[plucky elk]

it's a fact

[clever karma]

OHH

okay

yes

i know that

[plucky elk]

do you think there's a way you can add two positive numbers to get a negative?

[clever karma]

no

[plucky elk]

right

what do you think the correct answer to "solve" would be then?

[clever karma]

to make the exponents negative?

im not sure

[plucky elk]

if there's no way to add two positive numbers to get a negative one, but someone asked you for them, what should you tell them

[clever karma]

thats not possible

[plucky elk]

right

[clever karma]

😎 😎

[plucky elk]

that's what you should put

[clever karma]

oh wait

so its no solution

[carmine tulip]

What about complex solutions?

[plucky elk]

yes no solution is another way to put it

[clever karma]

OMG IT IS HAHAHAHAHHAHA

[clever karma]

im not that highyet

[carmine tulip]

Does the equation have any complex solutions?

[plucky elk]

read #help-23 message

[carmine tulip]

okay but I'm asking generally.

[plucky elk]

#❓how-to-get-help

[carmine tulip]

bruh

[clever karma]

LMAO

wait for 3 times 4^2x + 5 times 4^4x=2

do i combine the times first

btw the times are like the little dots

[plucky elk]

you can't combine them in any meaningful way

you just use substitution

[covert yoke]

$3 \cdot 4^{2x} + 5 \cdot 4^{4x} = 2$

[flat frigate]

OmnipotentEntity

[covert yoke]

@clever karma this?

[clever karma]

si si

wait no

the 4^x

is just to x

[covert yoke]

if so what happens if we replace u = 4^(2x)?

[clever karma]

not 4x

[plucky elk]

you typo'd here

[covert yoke]

oh

in that case let u = 4^x

[clever karma]

im sorry guys

omg wait

so u make it a quadratic?

[covert yoke]

yes!

[clever karma]

so whenever i see a common base in a problem i use substitutinm.>?

[covert yoke]

you can try it

[clever karma]

wait so when i have 3 dot u^2 + 5 dot u=2

what do i do?????????

just combine?

[covert yoke]

You know how to solve a quadratic yes?

[clever karma]

yes

[covert yoke]

just do that

[clever karma]

wait even if it has a dot

can i just seperate the 3 dot u^2?

[covert yoke]

the dot is just a multiplication

[clever karma]

si

[covert yoke]

3 dot u^2 is just 3u^2

[clever karma]

oohokokk

so 3u^2+5u-2=0

and x cant b negtivr right

like in the context of the question how do i check?

do i plug it in for x?

[covert yoke]

x can be negative, but u is the result of an exponentiation, so it can never be negative (unless we count complex results.)

[clever karma]

so my solutions are -2, 1/3

but it cant be -2

like what do i plug that in to?

[covert yoke]

u = -2 and u = 1/3?

[clever karma]

every x in the og equation?

si

[covert yoke]

well, u = 4^x

so you have -2 = 4^x and 1/3 = 4^x

The first has no real solutions

the second you can get by using logarithms.

[clever karma]

pojhh i jeep forgetting that i have to check with that u equals

okayyy tysmm!!

.close

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[clever karma]

.open

.reopen

[safe radish]

✅ Original question: #help-23 message

[clever karma]

3ln(6x−3)−5=16

[visual linden]

okay simplify it down into ln(f(x)) = C

[clever karma]

😭 😭 idk hiw

putting the 3 as a power?

[visual linden]

no like just add 5 on both sides

[clever karma]

oH FUH

okay i got that

what next?

[visual linden]

now divide 3 on both sides

[clever karma]

okay

ln(6x-3)=7?

[visual linden]

yes

[clever karma]

do i raise to e?

[visual linden]

now raise both sides to the power of e

[clever karma]

to get rid of ln?

[visual linden]

yes

[clever karma]

omggg tysmm!

muchos gracias

.close

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[leaden knot]

best way to improve mental math?

[leaden knot]

And how many minutes to complete this kind of mental test would be considered average or good?

[summer coral]

best way to improve on mental math - practice practice practice. know what methods to use like either breaking it down, rounding, memorizing common facts like 1/8 = 0.125, 0.5 = 1/2, etc.

[leaden knot]

best ways to practice?

[summer coral]

if you want to go for speed maybe try to go for < 30 sec per problem, if you cant do that right off the bat go for < 1 min, set your own goals and keep beating them.

[summer coral]

literally just find problems on google and well... do them

[leaden knot]

thanks!

[summer coral]

ofc anytime

[safe radish]

If you are done with this channel, please mark your problem as solved by typing .close

[marsh gorge]

.close

[west bramble]

practice over and over again

its really easy

just practice

be patient

[leaden knot]

ty

.close

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[golden fog]

help with question, please

[safe radish]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[golden fog]

I have begun but got stuck midway.

[gusty flicker]

What is it asking to find

I don’t speak Portuguese

I think it’s in Portuguese at least

<@&268886789983436800> …

[golden fog]

sorry

[gusty flicker]

Nah I pinged the mods about a scammer

Not u

They deleted the message lol

[golden fog]

ok

I resolved the issue, thank you.

[gusty flicker]

Oh ok

[safe radish]

@golden fog Has your question been resolved?

[safe radish]

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[pallid condor]

lemee rewrite this

Barney is an ecologist who studies the change in the tiger population of Siberia over time.
The relationship between the elapsed time
t in decades, since Barney started studying the population, and the number of tigers,
N(t) is modeled by the following function:

[flat frigate]

ΡΙ@nnΣdVφdhMDCXXI

[pallid condor]

the tiger population loses 3/5 of its size every _________ decades

[safe radish]

Channel closed

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[pallid condor]

no wait

bro

[quiet plume]

Just open another

[pallid condor]

k thx

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[cloud hill]

sqrt(1+2*sqrt(1+3*sqrt(1+4*sqrt(1+5*sqrt(...)))))=?

[cloud hill]

friend sent it to me

i have no idea what's the right approach

i've tried looking at functions whose inputs somehow generate that

like this

manipulating the expression itself doesn't work

at least not to my knowledge

and i don't know how do define it as a sequence whose limit i could take

i have to get off now, if this question doesn't get any responses someone close it please

[safe radish]

@cloud hill Has your question been resolved?

[tardy mango]

you can google this

Ex. see the comment on this post

Basically be really clever in rewriting things as perfect squares

[cloud hill]

ah okay

i was at least right in the fact that some sort of variable was needed

but idk if i would have thought of that

ty

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[steady kestrel]

Working on the following problem in an algebraic geometry class. I see how to induce the homomorphism, but showing that it is injective and surjective eludes me. I'm unsure how to proceed, especially in the surjective case

[plucky elk]

Try uhhhh #algebraic-geometry

#get-advanced-access if you need it

[steady kestrel]

sure

.close

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[night heath]

I remember the method had something to do with using nCr

[mossy lotus]

yep, its just choosing 2 people from amongst the n to shake hands

[night heath]

how do you express nCr in factorial form again?

[vague solar]

seein things like these really disappoint me on what they teach us rn 🥀 💔

if only we could be taught such wonderful questions actually concerning maths and not simply rote learning ah.

[night heath]

ive been scrolling all day and I have a test on this in less than 9 hours

[vague solar]

😭 , that sucks man. I am just done with my maths test with the most lengthiest paper in history god dang it. Got my english test tmrw and here i am in the mathematics server xD.

[solar hazel]

we have ${n \choose 2} = \frac{n!}{(n-2)!2!}$

[flat frigate]

slayla

[solar hazel]

a little bit of algebra will get you there

[night heath]

n(n-1)/2

yea that works out

can someone explain Binomial expansion with the example of $(3a-b^2)^9$ (Ik pascals triangle so dont bother w/ that

[flat frigate]

Cowking

[night heath]

like help me find the first 3 terms

nvm

.close

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[fringe dock]

for part e the answer is that it has a staircase pattern and thereforeit converges

but earlier he says that if it has a staircase pattern it diverges

i am really confused

I don't really understand it either

i just know how to draw the pattern on the graph

[median vigil]

(a) is a convergent staircase, (c) is a divergent staircase

[fringe dock]

so (a) bounces to go towards the intersection and (c) bounces to go away the intersection?

[median vigil]

exactly

[fringe dock]

how about (b) and (d)

do i just

remember that b gets closer and closer

to the intersection

and d doesnt get closer and closer

[median vigil]

yeah

[fringe dock]

is that all

[median vigil]

the whole idea of convergence is that each iteration brings you closer to the intersection, and divergence is that it does not do that

[fringe dock]

ohhh

okay

one thing

what does it mean when i find something that is closer or further away from the intersection?

I am not sure what i'm doing other than following a pattern I've memorised

[safe radish]

@fringe dock Has your question been resolved?

[safe radish]

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[woven fjord]

Heres my working i dont know how to continue the second sum

[safe radish]

@woven fjord Has your question been resolved?

[safe radish]

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[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sinful token]

ok guys

[sinful token]

how to be good with the redaction in math

any tips

[ebon igloo]

what even is this ?

[drifting sierra]

Redaction?

[sinful token]

ERM

writting

i meant

writting math

[ebon igloo]

what standard

[drifting sierra]

Writting

[sinful token]

ensure that the writing is attractive and rigorous

[drifting sierra]

Right

[sinful token]

without spreading yourself too thin and ensuring that the formulas are structured

[ebon igloo]

so you want to learn how to write mathematics?

[sinful token]

guys this is a real problem 😭

yeah improve writing

[ebon igloo]

what standard?

[sinful token]

stoop the crying emoji

😭

[ebon igloo]

i cant stop

[sinful token]

standard?

[ebon igloo]

like

college or what?

[sinful token]

yea college

[ebon igloo]

What a good question

[sinful token]

i agrree w emoji tht time bro

[sinful token]

WHAT

[ebon igloo]

how to write maths

[sinful token]

yeah

any tips guys

[ebon igloo]

hmm

[ebon igloo]

move your hand slowly🥀

[honest perch]

can you write greek letters?

[sinful token]

some yea

[honest perch]

good

[ebon igloo]

like alpha beta gamma ?

[sinful token]

yeah lambda ect

but only that

[honest perch]

if you want to prove something that start with "for all eps>0 ..." then in ur proof it's good to start with "let eps>0"

[ebon igloo]

brother ....she wants tips on how to make her writing in maths good

[honest perch]

? yes i know

[ebon igloo]

[sinful token]

what is funny 😭 ??

is it a help channel or what

[honest perch]

nothing

is funny.

[ebon igloo]

now how to help you in that

[sinful token]

thats why i dont find any good videos on that

[ebon igloo]

like the only good thing i can say is write better

[sinful token]

is it a sensitive subject orr

[ebon igloo]

maths

[sinful token]

yea?

[calm cloak]

Write numbers 1-100 daily two times

[ebon igloo]

you are the first person who ever got this problem

[sinful token]

pretty sure im not thhe only one!!

[sinful token]

Whht why :DD

[ebon igloo]

hm

[ebon igloo]

thats the best way

[obsidian ginkgo]

Learn what different symbols mean, there's different symbols that mean for all, since etc
Also read formal proofs and write how they write

[sinful token]

alright!!! thx

[honest perch]

if ur thinking is clear, ur writing will also be clear

[sinful token]

I see

[honest perch]

this is kind of an example of that

like u wanna fix eps>0 in the beginning

and u don't change it for the rest of the proof

you could also write "fix eps>0"

[sinful token]

Ohh i tend to forgot that tbf so thanks

[obsidian ginkgo]

Also checkout a guide to writing math by Dr Kevin Lee

[proud tree]

just a general tip for stuff like proofs: avoid symbol soup, words are much easier to understand but I'm not telling you to not use symbols at all

[honest perch]

also if ur defining a new symbol (f) like f = g - h, then make it clear that ur defining it and it doesn't follow from somewhere else

like say, let f = g-h

instead of just f = g-h

[sinful token]

sometime we need to use like it is an obligation but i do understand you mean

[sinful token]

i just found it thanks! im going to watch that :))

[sinful token]

Do you think it's a good idea to give the definition before describing the entire demonstration process?

I would like to dothat

[honest perch]

not necessarily. define things in ur proof where they naturally arise

eg if u wanna define delta_1 = eps/2 and delta_2 = eps/3. do that where it should occur most naturally, not necessarily at the beginning

[deep flint]

?

[white violet]

f := g-h

[honest perch]

also valid

[proud tree]

as to what content is apt to write, it honestly depends, for example if you are taking a particular course, you might use definitions and theorems almost without mention if you're deep into it but would probably be particular about them earlier in

[sinful token]

a definition with words i mean yeah like a property

[honest perch]

are u writing a book

[sinful token]

💀

yall should see the teacher

[honest perch]

no i mean, what are u writing

or u just want general advice

[sinful token]

general advice for answering exercice

or test

[honest perch]

is it proof based?

[sinful token]

nope

they dont do it yet

im a computer science student

but we have some math

[honest perch]

oh 💀

[sinful token]

thanks God i am not into that

😹