The Grove

Prince North Læraðr

20:21

@bobble Tree requests more help

bobble

crown here

Prince North Læraðr

I'm going to fail this test on Friday ;-;

bobble

tree will not fail test

tree must climb out of depression-pit

Prince North Læraðr

bobble

for #3, what have you done so far?

Prince North Læraðr

20:23

I've managed to reduce the equation down to 2x^2+x=e

or I guess x(2x+1)=e

bobble

i see a quadratic

Prince North Læraðr

It doesn't work in the quadratic formula....

bobble

why not?

Prince North Læraðr

Oh wait it's minus e

Gross

x=[-1+sqrt(8e+1)]/2, [-1-sqrt(8e+1)]/2

bobble

solutions!

Prince North Læraðr

20:27

Okay so for number four I've simplified down to

5-log_x(9)=log_3(x) or 5=log_3(x)+log_x(9)

bobble

hmm bobble has broken brain

could try "absorbing" the 5 into a log so you just have a log on both sides?

Prince North Læraðr

The 5 was from log_x(x^5)

bobble

stick the 5 into the other log?

Prince North Læraðr

Like log_3(3^5)?

bobble

flailing bobble noises

Prince North Læraðr

20:34

Shall we summon deus?

bobble

start the ritual

Prince North Læraðr

@Deusovi We request thine presence

We must now uh

Start making grid-deductions and CCs

Throw physics into the air until he comes

(Well, while we wait for Deus, I'm gonna just move on and work on other question)

Is there a way to undo log operations?

Like

I have this equation

logx^2=log2+log(x+4)

logx^2=log(2*x+4)

logx^2=log(2x+8)

I would like to get rid of the logs so I can do quadratics

I mean I know in a fundamental level I can just ignore the logs because if the insides are equal then the equation is true

bobble

If I throw a packet of math homework (m=0.5kg) at a log (m=100kg), at an angle θ=30 degrees, with initial velocity v=10m/s, by how much will the log accelerate at the moment the math homework hits?

hehe

anyways, to your problem

Prince North Læraðr

But I just need like an operation

Is it log^-1? ARCLOG

bobble

what's the opposite of a log?

Prince North Læraðr

20:42

I'm dying here

ARCLOG

bobble

log is inverse of EXPONENT

Prince North Læraðr

Sorry

No, ARCLOG

bobble

do 10^ both sides

Prince North Læraðr

I'm kidding, yes it's exponent

bobble

I'm confused at what you were trying to convey

Sid

20:43

@PrinceNorthLæraðr yuck.

Prince North Læraðr

What do you mean?

@bobble No because like inverse sine is arcsine, and it's written as sin^-1

I was being facetious

bobble

yeah, see I didn't get that

Prince North Læraðr

Ah

bobble

I thought you were legitimately asking "what would the equivalent of an arclog be?"

Prince North Læraðr

Oh, no, I was messing around

bobble

20:44

and "I'm dying here" meant "I'm struggling and I can't figure out what to do"

Prince North Læraðr

Ah, no

Sid

I am fairly sure there is some sort of logarithmic property involved in your solution.

Prince North Læraðr

Hm, can x equal both -2 and 4 in the question that I just posted?

because 2log(x)=log(x^2)

So 2log(-2) is log((-2)^2)

Sid

@PrinceNorthLæraðr log of a negative number is not defined.

At least not for real numbers, from what I recall

Prince North Læraðr

Hm, but the 2 in front of the log....

Because it just becomes log(4)...?

I do get what you're saying though

Sid

20:48

@PrinceNorthLæraðr yeah sure but that's log 4. It's not 2 log (-2) because log(-2) is undefined

You can't multiply 2 to an undefined number and expect to get a reasonable answer

Prince North Læraðr

Hm, so I can't do 2log(negative number) despite the fact that it's log(negative number squared)

Interesting

Sid

@PrinceNorthLæraðr yeah. This is tedious imo. You have to expand the RHS and then assume log x(base 10) to be y and solve a quadratic equation

Prince North Læraðr

Yikes. I'll do that in a bit

Deusovi

22:05

@PrinceNorthLæraðr hello! i'm around

Prince North Læraðr

(okay, hold on :)

So we have this thing right here

And we're kind of stuck on number three

Deusovi

22:21

hm, ok

first move all of them to one side, and then consider factoring something out - you should notice A Thing happening if you can

Prince North Læraðr

Okay

Oh wait no this is the wrong question

It was number 4

My bad

(I was going through it and I realized oh wait, I just did this)

We've gotten it down to 5-log_x(9)=log_3(x)

Deusovi

oh

do you know the change of base rule for logarithms

Prince North Læraðr

Yes

If I have log_y(x), then it's ln(x)/ln(y) (using natural log as an example, but I can use any base that's not x or y)

Deusovi

"I can use any base that's not x or y" - why not x or y?

Prince North Læraðr

Oh... I just assumed you couldn't because that would be redundant?

Deusovi

22:34

you can - it might not be useful, but redundant statements aren't banned in math

lots of clever tricks for solving problems are just "multiplying by 1" or "adding 0"

Prince North Læraðr

Well not y, because that just gives you the same expression

Hm

Deusovi

but i wouldn't use x and y in your expression here, because you have an x in your problem that might take different roles

anyway

i don't like x being in the base of logₓ(9), so i would try to fix that first

see if it makes things easier to deal with

Prince North Læraðr

Okay

We get 5-(ln(9)/ln(x))

And then ooh do the same to the other side

Oh, I thought I had something there with cancelling stuff out but anyways

Deusovi

hm

i'd use log₃ rather than ln

Prince North Læraðr

Okay

Deusovi

22:38

because you already have log₃ in your problem and that might let you do some nice combining

(you could also do the same to the other side, yeah, but that sounds like more work)

Prince North Læraðr

(okay)

Hm, could multiply everything by 3^? That might work

Deusovi

"3^" is not a number

what does it currently look like?

Prince North Læraðr

I meant raise the expression by "3^whatever"

5-ln9=log_3(x)*ln(x)

Moved the x's to one side

Deusovi

why ln?

why not log_3, since you already have that?

Prince North Læraðr

oh, that's just the base I used for change of base

Oh, true

3=(log_3(x))^2

Deusovi

22:43

oh this looks much nicer

Prince North Læraðr

Because 5-(log_3(9)/log_3(x))=log_3(x)

Deusovi

yep!

Prince North Læraðr

Move stuff around and simplify

Hm, I can express 3 in terms of log_3 to get rid of log_3

Deusovi

can you? not sure you can get rid of log_3 that easily

because it's hidden under a square on the right side

Prince North Læraðr

Well

log_3(27)=log_3(x)*log_3(x)

Oh I see the problem

Deusovi

22:45

yeah raising 3 to both sides just gives you something gross

@PrinceNorthLæraðr from here i think there's a pretty easy thing you can do

Prince North Læraðr

U subsitution, maybe?

log_3(x)=y

y^2=3

log_3(x)=sqrt(3)

Deusovi

i would've just taken the square root of both sides, but that's perfectly fine

hold on a sec

HTM

@PrinceNorthLæraðr I don't think that simplifies down to what you have right now

Prince North Læraðr

Wait nope

HTM

The "5 - " isn't part of the numerator of that fraction

Deusovi

22:47

ah yeah i agree with HTM

Prince North Læraðr

Oh, you're right

HTM

I would recommend going back to the original problem and taking log base 3 of both sides

Deusovi

no i think it's fine to continue from that one expression

Prince North Læraðr

Oh, I was going to multiply everything by log_3(x)

And then quadratic

Deusovi

yeah that sounds good

HTM

22:49

@PrinceNorthLæraðr Yeah, that's what's intended

Deusovi

(i think log₃ of both sides in the original problem would do roughly the same thing, but it's easier to see from the current expression)

HTM

You automatically get (log_3(x))^2 when you take log_3 of the original equation

Anyway, you get to the same quadratic no matter what route you take

Prince North Læraðr

Ew this is a gross value for our "u"

HTM

@PrinceNorthLæraðr Then you're on the right track

Prince North Læraðr

log_3(x)=(5+sqrt(13))/2, log_3(x)=(5-sqrt13/2)?

Is x like 3^((5+sqrt13)/2), 3^((5-sqrt13)/2)?

HTM

22:54

That's almost correct, check the radicand

Prince North Læraðr

radicand?

Deusovi

thing being rooted

HTM

The number under the square root

Prince North Læraðr

Ah okay

HTM

Also don't forget to plug the values back in to check that they work

Prince North Læraðr

22:56

Um 25-8=13 right?

OH I literally cannot math

How did

My mind is clearly non-functional

I went 5+8=13

HTM

@PrinceNorthLæraðr Hey don't worry, I originally have 25 - 8 = 19 for some reason

Prince North Læraðr

We're like

Missing all the odd numbers

sqrt(17)

Ugghhh

I hate this :(

bobble

did tree's brain take a vacation

HTM

@PrinceNorthLæraðr Well, it'll only get better with practice, I assure you

Prince North Læraðr

internal screaming

HTM

23:04

I know it's cliche to say that, but much of math is about recognizing patterns and finding ways to tackle hard problems, so practice will only give you more tools to do so

Prince North Læraðr

How do you simplify 4=x^2(-x+3)?

Well okay

Here's the question:

Deusovi

step 1: take log of both sides
step 2: hope that that helps

Prince North Læraðr

ln(4)+ln(x+3)=ln(x^2)+ln(9-x^2)

HTM

22 hours ago, by HTM

@PrinceNorthLæraðr Hint: if this is part of the review on logarithms, then that's probably what you should try first :)

Prince North Læraðr

ln(4*(x+3))=ln(x^2*[(-x+3)(x+3)])

raise it from e? by e? whatever

HTM

23:09

@PrinceNorthLæraðr I just looked at it, and that's what I have as well

Prince North Læraðr

natural logs go poof

Deusovi

oh, 4=x²·(-x+3), not 4=x^(2(-x+3))

Prince North Læraðr

Oh, my bad

Parenthesis

How do you simplify 4=(x^2)*(-x+3)?

Deusovi

~~cubic formula~~

Prince North Læraðr

Does this require me to do like long division or something

HTM

23:11

Have you expanded out the right hand side yet?

Prince North Læraðr

(-x^3)+(3x^2)-4=0

x^2=u? Idk

HTM

Aight, now you should be able to solve for x

Hint: find an "obvious" solution first

Prince North Læraðr

Uhhhh

Ig x=1 works...?

HTM

You've learned how to factor polynomials, right?

Prince North Læraðr

No

x=-1

@HTM Last semester. My brain is a bit nonfunctional right now

HTM

23:13

@PrinceNorthLæraðr Good time to review that then :)

Prince North Læraðr

uggghhhh

Oh there was that weird one um

With the box

HTM

OK, so if x = -1 is a solution to the equation, then what can you say about the polynomial -x^3 + 3x^2 - 4?

Specifically, what divides into -x^3 + 3x^2 - 4?

Prince North Læraðr

(x+1)

There was that

I forgot what it was called

HTM

Ye, now you can long divide to cancel out that factor

Prince North Læraðr

There was that shorthand variation

What was it called?

HTM

23:16

@PrinceNorthLæraðr You mean synthetic division?

Prince North Læraðr

Yessss that one

Synthetic division works here because x+1

and x is to the singular power

HTM

OK, do you remember how to do it?

Prince North Læraðr

Yeah

-1 is to the right, and then you make like this box

And then you do stuff

My brain's not working right now so I can't explain but I can do it

Okay so our terms remaining after synthetic are -1, 4, and -4

So -x^2+4x-4

That's ugly so -((x^2)-4x+4)

Which is just -((x-2)^2)

x=-1, x=2

internally screams

HTM

OK, now don't forget to check that these values actually work

Prince North Læraðr

Darn it

HTM

23:21

@PrinceNorthLæraðr Avi, is that you?

Prince North Læraðr

Hehe

Aw, I miss Avi

And his stream of conciousness

Okay -1 works

2 works

Both works

Deusovi

hooray!

HTM

:thumbs_up:

Prince North Læraðr

;-; I'm so scared for the test on Friday

Deusovi

you got this!

Prince North Læraðr

23:25

~~I hope whoever invented logarithm is suffering in his dead grave~~

Deusovi

oh boy do i have good news for you then

HTM

@PrinceNorthLæraðr Get revenge by using his bones

Napier's bones

Napier's bones is a manually-operated calculating device created by John Napier of Merchiston, Scotland for the calculation of products and quotients of numbers. The method was based on lattice multiplication, and also called 'rabdology', a word invented by Napier. Napier published his version in 1617. printed in Edinburgh, dedicated to his patron Alexander Seton. Using the multiplication tables embedded in the rods, multiplication can be reduced to addition operations and division to subtractions. Advanced use of the rods can extract square roots. Napier's bones are not the same as logarithms...

Prince North Læraðr

BWAHAHAHAHA

Deusovi

(aw my mind blanked on the name but i was looking for that)

Prince North Læraðr

I'm coming for you Napier

Your bones will be used for my fertilizer >:D

HTM

23:29

I mean, they're not literally bones

I actually have no idea why they're called bones

Deusovi

they're not literally bones if you're a coward

3

Prince North Læraðr

Bwahahaha

HTM

@Deusovi bruh

in The Sphinx's Lair, Apr 1 at 17:18, by Deusovi

well, that's ominous

Prince North Læraðr

Okay so we have all these dumb long problems and then the last problem of the section is this:

It's just doing "4^n" to both sides

North's brain has malfuctioned. Please hold while the system resets

Yes it's 4 to each side

Except you convert 4 into sqrt(2)^4 for the right hand side

And poof, 4*1/4=1, 2x-1=x+1

Really gotta do me like that, huh math

Deusovi

sounds reasonable to me!

√2 to each side would've also worked - there are also a bunch of other ways of doing it but they all end up being the same idea

Prince North Læraðr

23:35

Oh, I guess so

Except the right side would be sqrt(2)^4

And then it'd be 4(2x-1) and the expression would've resimplified down

Deusovi

yeah

Prince North Læraðr

I still have hyperbolic trig after this, and then an essay, and then APUSH

Ugh

HTM

Ok, I gotta go, hope the studying goes well!

Deusovi

see ya htm!

hm, i can try to help with hyperbolic trig but i am not very familiar with it

Prince North Læraðr

@HTM See ya, thanks for the help :)

I was complaining to my friend the other day about how I don't have a love life and then I realized I don't even have time for romantic relationships at this point

@Deusovi It's pretty fundamental hyperbolic trig stuff from what I remember from class. Just remembering that cos(x)-sin(x)=1, and then stuff like sinh(x)=(e^x-e^-x)/2 and what not

Hey, that looks familiar

Oh, is hyperbolic trig derived from Euler's formula too?

Deusovi

23:38

yep!

Prince North Læraðr

Except no imaginary numbers

Welp time to finish this work stuff first

Hm (5^x)/(3^x) is not (5/3)^x right

Just sanity check

Wait it is

Deusovi

it is

Prince North Læraðr

If x=2, then (5^2)/(3^2), (5*5)/(3*3) or (5/3)*(5/3) which is (5/3)^2

Deusovi

hooray, sanity check!

Prince North Læraðr

Double-checking I did this correctly

3^x=5^(x-1)

3^x=(5^x)/(5^1)

5=(5^x)/(3^x)

Deusovi

23:45

makes sense to me

Prince North Læraðr

ln5=ln[(5/3)^x]

Deusovi

answer looks like it's gonna be gross

hold on

Prince North Læraðr

ln5=x*ln(5/3)

Deusovi

oh ok the ^x should've been inside those brackets

Prince North Læraðr

Oh right

Waaait

Eww

Wait yes

It is x=ln(5)/ln(5/3)

It throws me off that you can just divide a log expression

Deusovi

23:47

yep, looks good to me

if you want you can even do reverse change-of-base on that to get the """"nice"""" log[5/3](5)

Prince North Læraðr

LMAO

Lovely

I think I did something wrong with 3^(x+1)=e^(x/2)

Oh, nevermind, we're good

I did some variable manipulation wrong and ended up cancelling x out the first time

Deusovi

oh glad you fixed that then

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