help-36

[terse sorrel]

1 = sin(x+arcsin2/root5)

[vague anchor]

Kounsi trigonometry hai bhai

Inverse kae

Bina karna hai

Bhai

[terse sorrel]

x = arccos2/root5 if x < pi/2

use

[terse sorrel]

its not some hifi inverse trigo

[bold zenith]

!nosols

[final saddle]

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.

[terse sorrel]

use general term frm here

[vague anchor]

What I've learnt is trigono and just it's basic halfway useless formulae

[terse sorrel]

5th is similar

[vague anchor]

X kae general solution chaiyae

[terse sorrel]

yeah

u get cosx = 2/5 in 1

[vague anchor]

To what

[terse sorrel]

to 1

[vague anchor]

F
I already solved it solve me 5th

!nosols

[final saddle]

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.

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.

[plucky rover]

Okay what point are we at

Have we made any progress at all

[vague anchor]

5-(i)

[plucky rover]

No I got that much

Show what you've done so far ig

[vague anchor]

I expressed the equation as quadratic in Tan x/2

[plucky rover]

Don't describe it, show it

[vague anchor]

Ohk so here t is tanx/2

Ignore the last thing

[plucky rover]

This feels like you're overcomplicating it

You could just express sin in terms of cos

And then get a quadratic in terms of cos x

My guy you want the text outside the dollars

And only the latex stuff inside the dollars

[bold zenith]

Struggling here with phone keyboard in uk mode without dollars lol

[plucky rover]

Lmao fair

What happened to the bot

Move $cos(\theta)$ to the right side of the equation and square it

[soft zealot]

Xavier 🌺

[plucky rover]

Spacing hmm

[bold zenith]

.-.

[plucky rover]

The space before the ending $

Also no that's still more complicated than necessary

No remove the spaces lol

Don't add more

[bold zenith]

I know, Move $cos(\theta)$ to the right side of the equation and square it

[soft zealot]

BBMaths

Xavier 🌺

Write $sin(\theta) = \sqrt{1-cos^2(\theta)}$

[bold zenith]

Don’t need that, if you just square you don’t need to worry about sqrt, but you must keep track of signs

[vague anchor]

Yepp wait

Ik what to do further

Calm

[plucky rover]

It's easier

[plucky rover]

Ah cool, good luck

[vague anchor]

I mean i still might need help later

[plucky rover]

Yeye feel free to come back

[bold zenith]

If you square both sides of an equation you have to double check the answers you might have fake solutions

[plucky rover]

Yes

That would be true for both of our methods lol

Either way, let him do it for now ig

[vague anchor]

Check it buss it buss it

@plucky rover

[plucky rover]

,rccw

[vague anchor]

Messy🤮

[plucky rover]

It's your writing

I genuinely do not understand how you got between line 2 and line 3

It looks like you cancelled out sin²x and cos²x on different sides

That's not how it works

[vague anchor]

RIP

[plucky rover]

Also genuinely do this, makes your life much easier

[vague anchor]

Then what to do with that sqrt

Squaring both side?

[plucky rover]

Yes

[bold zenith]

I personally prefer squaring first

You don’t need to define sqrt

[plucky rover]

It really doesn't matter tbh

[vague anchor]

I did that only above

[plucky rover]

You didn't substitute

Substitute first, then square

[plucky rover]

Also you did this

[vague anchor]

Isn't both same

[vague anchor]

Nahh i redo it

Silly mistake 😜

[plucky rover]

You know we can't read your mind right

[vague anchor]

Btw have u seen Ann??

[plucky rover]

Send what you have now

[plucky rover]

I do not see how that is relevant

[vague anchor]

Wdym

[plucky rover]

^

[vague anchor]

Oh wait im gonna write it properly first

,rccw

[plucky rover]

How did you do the last step

[vague anchor]

I mean switch +- and -

[plucky rover]

No I mean where did the 2 √2 come from

[vague anchor]

Oh man that makes 6 not 8 😅😭

[plucky rover]

Good observation

[vague anchor]

Btw what to do further i mean squaring it would have increased its roots

@plucky rover here?

Xavier

<@&286206848099549185>

[bold zenith]

When you have cos theta just arccos it

One of the solutions might be fake because you squared but the other you can now find

[vague anchor]

Ohk

.close

[final saddle]

Channel closed

Closed by @vague anchor

Use .reopen if this was a mistake.

[final saddle]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[radiant carbon]

bruh what did i do wrong

im so close i can feel it

[valid carbon]

!show

[final saddle]

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

[radiant carbon]

yeah wait opening up discord on my phone

[valid carbon]

,av x

[soft zealot]

Too many matching members found! Please refine your input and try again.

Sep 19, 2025, 5:23 PM

[valid carbon]

,av @radiant carbon

[soft zealot]

62pz's Avatar

Click here to view the image.

[valid carbon]

markov chain?

[radiant carbon]

can't remember, random diagram on my pc

[valid carbon]

what is $(9 \sin(\theta))^2$

[soft zealot]

Περσυ

[radiant carbon]

81sin^2 (theta)

[valid carbon]

yeah

so

[radiant carbon]

yeah i see it

so

would you sub in x = 3sin(th)

nah

that still gives wrong answer

[valid carbon]

yes

[valid carbon]

lol show

[radiant carbon]

💀 bruh

[valid carbon]

seems fine to me.

[radiant carbon]

im supposed to get this

[final tangle]

they're equivalent, do "something"
so that you won't have ugly fractions under the root

[radiant carbon]

how tho

i can't just manipulate the numbers under the root

without manipulating the rest

numbers under root need to be x9, but denominator needs to be x3

[final tangle]

consider multiplying numerator and denominator by 3

or factoring sqrt(1/9) out of the root

[radiant carbon]

bruh

im stupid

nvm

thanks i didnt see that

.close

[final saddle]

Channel closed

Closed by @radiant carbon

Use .reopen if this was a mistake.

[final saddle]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[final maple]

i got help from a TA for this question

thats the right answer

i understand everything

up until the part where i put the star

can anyone help me understand what exactly happened?

why was 2^7 broken up?

[pliant shore]

cause in the line after the star, you can combine the like terms

if you let $\sqrt[6] {2}$ be $x$

[soft zealot]

south

[pliant shore]

you have $x - 2 \cdot x = x - 2x$ (you made a mistake when writing that line)

[soft zealot]

south

[pliant shore]

it's not 2^6 * sqrt(2)

[final maple]

so he was trying to make the like term be: 6root2?

[pliant shore]

$\sqrt[6] {2 \cdot 2^6} = \sqrt[6] 2 \cdot \sqrt[6] {2^6}$, do you agree?

[soft zealot]

south

[final maple]

wouldnt it be root 6 then 2^7?

[pliant shore]

$\sqrt[6] {2 \cdot 2^6} = \sqrt[6] {2^7}$ is true, yes

[soft zealot]

south

[final maple]

okok

so where was the mistake

cuz i saw he broke up root6 2^7

[pliant shore]

$\sqrt[6] {2 \cdot 2^6} = \sqrt[6] 2 \cdot \sqrt[6] {2^6} = \sqrt[6] 2 \cdot \textbf{2}$

[soft zealot]

south

[final maple]

so

I understand the root6 (2), but i dont understand how you got another root6

[pliant shore]

yeah I think you misread your own work

[final maple]

this was the TA

it wasnt me

[pliant shore]

oh ok, let me address this then

are you familiar with $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$?

[soft zealot]

south

[final maple]

radicals arent my strong suit, but that looks like it makes sense

[pliant shore]

it's a consequence of the law of indices

[final maple]

ive never seen it before but yes it makess sense

[pliant shore]

$(ab)^{1/2} = a^{1/2} \cdot b^{1/2}$

[soft zealot]

south

[pliant shore]

that's the same thing

[final maple]

okok

[pliant shore]

since this works if you replace 1/2 with 1/6 everywhere

[pliant shore]

then you can split the radical into two radicals, like so

each radical is the 6th root

[final maple]

yes i see

i understand that

[final maple]

so where the star is

you mentioned like terms

you said the TA messed up

how so

cuz like

i understand yea he broke up 2^7

but the 2^6 js moved to the front and its exponent was gone

[pliant shore]

I misread what they wrote

yeah $\sqrt[6] {2} - 2 \cdot \sqrt[6] {2}$ is much clearer

[soft zealot]

south

[pliant shore]

otherwise you might think there's a $2^6$

[soft zealot]

south

[final maple]

yea

i know he has that

[pliant shore]

okay, let's address the exponent was gone part

[final maple]

yes pls

[pliant shore]

it follows from $\sqrt[6] {2^6} = 2$

[soft zealot]

south

[pliant shore]

cause $\sqrt[6] {2^6}= (2^6)^{1/6}$, by definition of $\sqrt[6] {x} = x^{1/6}$

[soft zealot]

south

[pliant shore]

and now what's (2^6) ^ (1/6)?

[final maple]

2

[pliant shore]

right!

so that's how

[final maple]

ok let me try and absorb this

gimme a sec

actually another thing

what is the point of breaking up 2^7

is it so you can have root 6?

making them like terms

[pliant shore]

yep!

[final maple]

my test is next week

in the event i have a question similar to this

once i get to the line with the star

can i just do this operation

and put the answer in front of

this

[final saddle]

@final maple Has your question been resolved?

[final maple]

@pliant shore is this correct?

[past lion]

hi!

idk if it helps there

but u can put $$2^{6/2} $$ in factor !

mb i read ur paper wrong

factorise by

factorise by $2^{6/2}$, then u got $2^{6/2} [1-2]=-1 \times 2^{6/2} $

[soft zealot]

croustibest

[final maple]

that looks a little complicated

is what i asked prior to pinging south correct? would that work on a test

[past lion]

well it does work

so i believe yea

it's all about factorisation

are you cool with this notion?

[past lion]

and yea absolutely, you have to!

the squared root is just $\sqrt[6]{x}= x^{6/2}= (\sqrt{3})^6$

[soft zealot]

croustibest

[ornate tree]

@final maple Do you still need help?

[final saddle]

@final maple Has your question been resolved?

[final maple]

okok thanks

i think i got everything

.close

[final saddle]

Channel closed

Closed by @final maple

Use .reopen if this was a mistake.

[final saddle]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[analog galleon]

hi could I get some help with this

[sour zealot]

which step gives you trouble?

[shell condor]

What do you need help with?

[bold zenith]

!status

[final saddle]

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

[analog galleon]

1

[shell condor]

Well the solution's there, you don't get it right?

[analog galleon]

mostly like step 3

[analog galleon]

yeah

when it turns into 4 square root 16 i get lost

[shell condor]

Well that's fine

[analog galleon]

okay

im studying for the aleks so i just clicked show solution

[shell condor]

Well it was 16^(5/4), you can write it as (16^5)^1/4 or (16^1/4)^5

[stoic temple]

It doesn't mean 4 squareroot
It is the "fourth root"(I call it fourth root) of 16

[analog galleon]

as in?

[severe canyon]

What is 2⁴?

[analog galleon]

16 times itself 4 times?

[shell condor]

Yeah, basically 16 raised to the power 1/4

[analog galleon]

64?

[shell condor]

Answer this

[analog galleon]

oh wait

i did it wrong

its 32

[severe canyon]

No no (very) wrong

[analog galleon]

i multipled it by the 4

and then by the same numbers

[shell condor]

Ahhhh it's incorrect, try again

[severe canyon]

How did you get it?

[analog galleon]

not 2

[analog galleon]

2x2 = 4 4x2

an dso on

i see my mistake

8x2 = 16 16x2=32

stopping 4 times cause its ^4 right

[sour zealot]

that's 5x though

[analog galleon]

o

[severe canyon]

Exactly

[shell condor]

what's 2^3?

[sour zealot]

2^1 is 2, remember

[analog galleon]

16

[analog galleon]

ooo

[shell condor]

show the process

[analog galleon]

sorry we havent talked about this in college algebra

briefly went over it

in alg 2

[analog galleon]

2x2, 4x2, 8x2

[shell condor]

Basically, powers are multiplying the same number to itself a certain number of times

[analog galleon]

wait 2^1 isnt 4?

[sour zealot]

no

[analog galleon]

wouldnt it be multiplying by 2

itself

[analog galleon]

okay

[shell condor]

so 2^3 would be 2.2.2 (. is *), not what you mistakenly did

[sour zealot]

yeah you have 2x2, which is 2 2s

[analog galleon]

ah ^1 is one 2

i see

i feel dumb for not knowing this

[shell condor]

You're fine

so now, what is 2^3?

[analog galleon]

its always the most basic concepts i struggle with

[analog galleon]

2x2 = 4 so 8

[shell condor]

Perfect, and what would 2^4 be?

[analog galleon]

16!

[shell condor]

Great!

[analog galleon]

cause its multiplied by 2 again

alright so what does the 4 square 16 mean again

square root of 16 to ^4?

[shell condor]

Now root is basically the finding the number be multiplied x times to get the number we have

[shell condor]

No no

[analog galleon]

ah

[shell condor]

okay, so if we have x^2 = y, what would x be?

[analog galleon]

is y 16

x would be 4 if thats the case

[shell condor]

It's just an example

[analog galleon]

oh uh

idk

[shell condor]

Okay, so square root of y right?

[analog galleon]

oh yeah

cause you have ot get rid of the square

to move on

[shell condor]

Exactly

so 4 is square of 2, and 2 is square root of 4 right?

[analog galleon]

yes

so when x is on the upper outside of a root

its the root of that number?

so in this case 4 on the outside means sq of 4

= 2

then how many times that need to be multiplied by itself to get the number in the square root?

[shell condor]

Not exactly, the number outside the root sign indicated how many times a number was multiplied to itself to get the number we have

[analog galleon]

okay

but now im confused

it says 2^5

[shell condor]

yes

[analog galleon]

2x2 =4 4x2 = 8 8x2 = 16

wouldnt it be 2^4

OHHH

wait

the answer in the parenthesis is 2

but then multiplied by the ^5

[shell condor]

exactly

now if we multiply 2 to itself 4 times, we get 16

[analog galleon]

so numer in the sq ^ whatever is on the outside is whats on the inside

[analog galleon]

yeah

[shell condor]

no no

[analog galleon]

oh

can we try one together

i have to get 5 right

or 3 in succession

[shell condor]

so, the if the number outside the root is 4 and the number inside is 16, we get 2

[analog galleon]

yes

cause 2^4 = 16

if the number in the root was 32

[shell condor]

Perfect

[analog galleon]

the outside would be ^5

[shell condor]

Exactly, to give 2

[analog galleon]

yess

its making some sense now

[shell condor]

Now let me give you a problem to test you

[analog galleon]

alright

[shell condor]

If the number outside root is 3, and inside root is 64, then what should we get?

[analog galleon]

a number to the ^3 = 64

so

[shell condor]

Exactly, yes!

[analog galleon]

uhh

i wanna say 4

but it ends up being 32^4

[shell condor]

Perfect!

[analog galleon]

oh wait im rigth

i keep forgetting to consider that ^1 is a thing

i think that 4x4 is the ^1 step

when its actually 4^2

[shell condor]

Your this illustration is exactly what I wanted to convey

[analog galleon]

ah ok

[shell condor]

Don't worry, you'll get that in time

[analog galleon]

i got the idea

now its just the execution

but now i need to get to that part

the question now is 32 ^ 3/5

[shell condor]

Yes, I'm honestly proud dude, you got it without me stating it

[analog galleon]

you kickstarted my brain

[shell condor]

A different question?

[analog galleon]

yeah

i gotta get 3 right in a row

to moveo n

and that one we did didnt count

[shell condor]

Alright, so you tell me, what do you do here?

[analog galleon]

break it apart to the base

so 32 ( 1/5 times 3)

[shell condor]

Good

[analog galleon]

then smush the big number and 1/5 together and make the multiplied number to an exponent

so

( 32 1/5 ) ^3

[shell condor]

Yes

[analog galleon]

im looking at my notebook rn

i havent memorized the steps yet

[shell condor]

That's fine, just practise these questions

[analog galleon]

then make the denominator the outside upper root

and the big number in the root

[shell condor]

Yes

[analog galleon]

and old multiplied number is the ^

so in this case 3

[shell condor]

Yes

[analog galleon]

is it easier to just move from step 1 to 4?

i feel like if you know the importance of each part it'll be quite easy

[shell condor]

Well you can if it's clear to you

[analog galleon]

alright

[shell condor]

Yes but some things just become understood over a period of time, like those steps

[analog galleon]

thats fair yeah

ok ok so so far we have

uh

[shell condor]

Just remember to practise

[analog galleon]

32 in the root

5 is the upper

so what ^5 = 32

[shell condor]

Great thinking!

[analog galleon]

that statement is really helpful

[shell condor]

IKR!

[analog galleon]

what to the ^ = inside of root basically

[analog galleon]

its so peak

[shell condor]

Yes, it's great

[analog galleon]

and outside number is always the numerator right?

outside ^

[shell condor]

Yes

[analog galleon]

ooo im cooking

alright lemme try

[shell condor]

You sure are

[analog galleon]

oh wait

i messed up at the end

i thought it was 96

cause i multiplied 32 by 3

[shell condor]

Ahhhh that's fine

[analog galleon]

but i need to put ^3 on the number i get from ^ = whats in the root

it was 8

since 2^3 = 8

[shell condor]

Bravo!

That's perfect!

[analog galleon]

alright this one is 8 4/3

[shell condor]

You're getting the gist of it my friend

[analog galleon]

without even thinking 8 is gonna be in the square

4 on the outside

[shell condor]

Yes

[analog galleon]

3 on the left upper

what ^3 = 8

its 2 right

then put 4 on the 2

2^4

so its 16?

YOO I GOT IT RIGHT

[shell condor]

Exactly

[analog galleon]

its so much easier once you know the steps

[shell condor]

You got it so quick too!

[shell condor]

IKR!

[analog galleon]

haha im like that

once i get the concept down my brain just zips thru it

[shell condor]

I hope it helps you on your journey

[analog galleon]

but with that comes with making easy mistakes

[analog galleon]

yeah ive got 90 modules to get through

trying to atleast

[shell condor]

Ahhh well that just comes down to practise, stay consistent and you'll be saying goodbye to them in no time

[analog galleon]

i need a 46 on the aleks

[analog galleon]

can you help me with log?

or do i need to make a new channel

[shell condor]

Sure mate

[analog galleon]

alright give me like 2 mins

[shell condor]

Nah here's fine I suppose

[analog galleon]

let me get 2 more right

[shell condor]

Sure, let me know once you do

[analog galleon]

also id like help with radians and stuff too

i got a question on that

didnt know what to do

[shell condor]

I'm right here

[analog galleon]

never heard of a radian till i saw it

[analog galleon]

okay one moment

gotta use bathroom brb

[shell condor]

Sure

[analog galleon]

can you explain in basic terms what a radian is

layman’s terms I think it is

[shell condor]

You know pi right?

[bold zenith]

A radian is a special angle such that the length of the curvy bit of a part of a circle (an arc) is the same as radius of the circle

It’s about 57 degrees iirc

Since the circumference of a circle is $2*\pir$ there are $2\pi $ radians in a full circle (360 degrees)

[shell condor]

Well this guy here has layed it pretty well

[soft zealot]

BBMaths

[shell condor]

Yeah like yk how the circumference of a circle is pi times the diameter (diameter is just 2r)

[analog galleon]

aight im back

[analog galleon]

yeha

3.14

and so on

[shell condor]

And the circumference of a circle is 2 pi times the radius

That basically means that if you take the radius and try to overlap the circle with it, it'd take you 6 times the radius, and .28 more of it to completely cover the circumference

[analog galleon]

hmm

im a little confused

lets start with the log then get into that

[shell condor]

Now a radian is basically 1 part of it, like just one radius is taken and you use it to overlap a part of the circumference, that is a radian

[shell condor]

Alright sure

[analog galleon]

ok so the question i got was

2^5 = 32

and it was like write it in log

[bold zenith]

(Imagine an equilateral triangle (all 3 sides the same length) and a slice of pie at the same time) the angle of the pie is the radian

[analog galleon]

ive went over this rearranging in alg 2 but i forgot

[shell condor]

Alright, so in log, it's kinda tricky

[analog galleon]

so cutting it in half?

[bold zenith]

The 3 sides would be the same length

But one is curved

[analog galleon]

how is it the same if its curved

[bold zenith]

The angle between the not 2 curved ones

[analog galleon]

wouldnt be shorter

or longer

[shell condor]

First thing you gotta know, log is used to find the power

[bold zenith]

Longer, yes but the angle wouldn’t be 60 degrees anymore

[analog galleon]

okay

[analog galleon]

so its not equal anymore?

or is the length the same but the angle isnt

[bold zenith]

Yes

[analog galleon]

ah

[bold zenith]

The other 2 don’t have angles well they do but they’re 90 degrees

[shell condor]

So we already know the answer in log, we use our initial number and our answer to find the power

[analog galleon]

thats weird

[analog galleon]

okay

so how do i put 2^5 = 32 in log?

[bold zenith]

In a quarter circle it has 3 “90 degree angles”

[shell condor]

So the base of the log (the one you write just after and at the bottom of log) is your initial number

[analog galleon]

what does a quarter circle look like

[shell condor]

And the value inside the log is the answer you had

[analog galleon]

what scenario is the base always 10

i remember my teacher saying that

is it if its not shown?

[shell condor]

That is a traditional log, it is used for functions that increase exponentially

Like the Richter scale

[analog galleon]

a

h

[analog galleon]

ok so thats not relevant yet

got it

[shell condor]

a stage 3 earthquake is 10^3 times greater than a stage 1 earthquake

So if it's not written, then the base of log is 10

Now, the base of log is the number we're raising to a power and the value in a log is the answer we'd get from other method, and we have to find power

How would we write it @analog galleon ?

[bold zenith]

This one is 1 radian, all 3 sides have same length and the angle between the green and blue straight edges is 1 radian

[analog galleon]

my b

had to do smthn

[analog galleon]

radian being the curved?

[bold zenith]

Radian is an angle so it’s the angle between green and blue

[analog galleon]

im not sure

guide is please

[analog galleon]

ah ok

[shell condor]

Alright

[analog galleon]

lets talk about radians later

[shell condor]

2^5 = 32

[analog galleon]

trying to learn 2 things is making my brain explode

[bold zenith]

That’s it i think for radians

[analog galleon]

alright thanks

[shell condor]

Now the number you're raising to a power is always the base

[analog galleon]

so 2 is the base

big number = base

[shell condor]

Yes

[analog galleon]

gotcha

[bold zenith]

Other than converting between degree and radians but don’t need that right now as we’re focusing on power log stuff now

[shell condor]

And the number you're going through power method is the value in a log

[analog galleon]

it wsa like 320 degree to radians

or smthn

[analog galleon]

what would that be in this case

[shell condor]

My apologies, I meant the answer you'd get from power method is the value in your log

[analog galleon]

o

whats the power method agai

again

wait is it 32 in this case

[shell condor]

2^5 = 32

[analog galleon]

2^5 = 32?

[shell condor]

Yes

[bold zenith]

Are we trying to define the log of a number here

[analog galleon]

so 32 log base 2 = 5?

[shell condor]

Perfect

[bold zenith]

1048576 log base 2 is 20

[shell condor]

Basically show it isnlog form

[analog galleon]

whaaaat the hell

[shell condor]

Yup, it's very efficient for finding powers

[bold zenith]

I know my powers of 2 up to 2^20

[analog galleon]

wait i put it in my calc and 32 log(2) doesnt = 5

am i able to get bases on a ti 30?

[shell condor]

No no, the number on the bottom (the base) is 2

[analog galleon]

yeah ik

[bold zenith]

So the way log is notated is not how we described it

[shell condor]

and the number inside the log would be 32

[analog galleon]

but on the calc it says (2)

[analog galleon]

so 32log

or would it be log(32

2log32

[bold zenith]

$log_{2}(32)$

[soft zealot]

BBMaths

[analog galleon]

ah

[shell condor]

Yeah this one

[analog galleon]

how do i show a base on my calculator

[bold zenith]

If it just says log it’s base 10

[analog galleon]

okay

how do i change the base

[bold zenith]

If it says ln its base e=2.71828…

[shell condor]

There should be an option of log that allows inputting base

[bold zenith]

You need log square box

Type 2 in square box

[analog galleon]

wdym

[bold zenith]

$log_{[]}(32)$

[shell condor]

Try putting an additional square bracket

[soft zealot]

BBMaths

[analog galleon]

ah

it deletes the whole log

if i try to delete 32

what is ln

i think thats what my teacher showed us

to get that

maybe

[bold zenith]

Ln is the natural logarithm you can use that if you want

[shell condor]

log with base e

[analog galleon]

okay

i dont think i have an option to change base on this

[bold zenith]

$log_{2}(32)=\frac{ln(32)}{ln(2)}$

[soft zealot]

BBMaths

[shell condor]

damn, yeah that works too, uses an identity

[analog galleon]

"You can find log4(9) by taking log of 9 (use the log key) and divide that by the log of 4"

found this on google

[bold zenith]

$log_{a}(b)=\frac{ln(b)}{ln(a)}$

[soft zealot]

BBMaths

[shell condor]

Also I guess base 10 would be fine too

[analog galleon]

in this case

ln 32/ ln 2

[bold zenith]

$log_{a}(b)=\frac{log(b)}{log(a)}$

[soft zealot]

BBMaths

[shell condor]

As base e would just cancel out

[analog galleon]

i meant log not ln

[shell condor]

However this is an identity, as of now just use it to verify your answer

[analog galleon]

ohhh yeah

i got 5

[shell condor]

Both work

[analog galleon]

dope

[bold zenith]

The difference between log of different bases is that it is a multiple of log of the other bases, it doesn’t matter if it’s log or ln as the fraction cancels it out

[shell condor]

You get the notation?

[analog galleon]

kinda

i needa drive home

ill be back in like 5-10 mins

[shell condor]

Sure mate

[analog galleon]

cheers

[bold zenith]

1 thing you need to know to solve this is $\log(a^{b})=b*\log(a)$, this works with any log, (log,ln,log2,…)

[shell condor]

Yup, you might need this later, remember this

Ping us when you're back @analog galleon

[soft zealot]

BBMaths

[final saddle]

@analog galleon Has your question been resolved?

[analog galleon]

ok im here

gimmie like 2 mins

[shell condor]

Sure mate

[bold zenith]

Here’s a list of all small powers $\4=2^{2}\
8=2^{3}\
9=3^{2}\
16=2^{4}\
25=5^{2}\
27=3^{2}\
32=2^{5}\
36=6^{2}$

[analog galleon]

okay

so now how do we translate that into log

[soft zealot]

BBMaths

[bold zenith]

Up to 36

Can you repost the question

[analog galleon]

2^5 = 32

[bold zenith]

16^(5/4) was it?

[analog galleon]

back btw

[analog galleon]

oh I already figured that out

[bold zenith]

I mean you worked it out

[analog galleon]

yeah

[bold zenith]

Logs aren’t required for that

[analog galleon]

yep

just need to learn how to rearrange

[shell condor]

Alright apologies, just came back

[analog galleon]

no worries man

[shell condor]

So basically, remember I told you about the richter scale

An earthquake of 3 on the scale is 100 times more massive than an earthquake of 1 on the scale

[analog galleon]

eh

thats confusing

[shell condor]

So basically, 1 is 10
2 is 100
3 is 1000
4 is 10000
5 is 100000
Etc

[analog galleon]

alrifht

[shell condor]

See a pattern?

[analog galleon]

alright

yeah

10

[shell condor]

Yup

Now to make things easier, I'll make a correction

[analog galleon]

ok

[shell condor]

Alright I've made it

So basically 1 is 10^1

Scale 1, now remember, in logs, our answer is always the power

[analog galleon]

okay

[shell condor]

Urghhhh, wait just get it like this, the base is your number in the bracket, the number in log is your answer you get from power method, and the power is what you're finding

$log_{a}(b) = c$

[analog galleon]

uh

[soft zealot]

doctorstrangejr

[analog galleon]

ngl can you say big number little number

word it like that

[shell condor]

Can be written as a^c = b

[shell condor]

Alright sure

[analog galleon]

in this example 2^5 =32

[shell condor]

So big number goes at the base, little number goes in the log and the power is your answer

[shell condor]

Big number = a
Little number = b
Power = c

[analog galleon]

okay

so upper goes on other side of equal

[shell condor]

Yes

[analog galleon]

right side becomes little nukber?

number?

or big

[shell condor]

Wait, let's rephrase

Big number is 2

Power is 5

And let's call 32 the answer, okay?

Big number = 2
Power = 5
Answer = 32
That'll help you if you get confused in the future

So, now tell me again?

[analog galleon]

okay

[analog galleon]

tell you what

[shell condor]

How you'll arrange them in log

Like you said power goes on other side of bracket

$log_{a}(b) = c$|||| = ||||$a^c = b$

[soft zealot]

doctorstrangejr

[shell condor]

Get it?

[analog galleon]

uhhh a little

Ok so

in 2^5 = 32

[shell condor]

(The |||| are just for separating it)

[analog galleon]

it’s

log base 32 2 =5

?

[shell condor]

No no, remember your big number always goes at the base

[analog galleon]

yeah 32 is base

log base 32

[shell condor]

32 is the answer you got

[analog galleon]

oh

2 is base?

[shell condor]

Yes

[analog galleon]

ah

so log base 2 35

32*

= 5

[shell condor]

Just write this down somewhere

[shell condor]

Exactly

[analog galleon]

okay

let’s try some examples?

[shell condor]

That's what I was gonna say too

[analog galleon]

alright

this is what I’ve got so far

[shell condor]

Yes, that's perfect

[analog galleon]

okay

practice time

[bold zenith]

The key point of logs is $log(a)+log(b)=log(a*b)$

[soft zealot]

BBMaths

[shell condor]

Yeah another identity

[bold zenith]

More important one

[shell condor]

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

[soft zealot]

doctorstrangejr

[shell condor]

This one too

[bold zenith]

Logs turn multiplication into addition

[shell condor]

But don't focus on these just yet, these are identities you'll need ahead, but let's clear your basic for now

[bold zenith]

What are we solving here

[shell condor]

Just practising converting power form to log form

[analog galleon]

so log(8)+ log(4) is log(8x4)

[shell condor]

Yup

[analog galleon]

so just log(32)?

[bold zenith]

Yes, because log(8) is 3log(2) and log(4) is 2log(2)

And log(32) is 5log(2)

[shell condor]

Yup, that came from the power identity

[bold zenith]

$\frac{a}{b}=a*(b)^{-1}$

[soft zealot]

BBMaths

[bold zenith]

Another equation i like, with this you don’t need division ever again, but don’t do this teachers hate it or at least mine did

[shell condor]

Yeah it helps ngl

[analog galleon]

does this work for any fraction

[bold zenith]

Yes even b=0

[analog galleon]

so 4/3 = 4(3) ^ -1

[shell condor]

Yup

[bold zenith]

Yeah

Thing is 3^-1 is just 1/3 lol

[bold zenith]

That’s how you can show this

[shell condor]

Yeah, these are in exponents

[analog galleon]

yeah anything ^ -1 is just 1 over it

give me a log thing

to rearrange

[shell condor]

Alright

5^3 = 125

[analog galleon]

ok big numb is sub

so

[shell condor]

For the initial 1-2, you can refer the notes

[bold zenith]

log(6)+log(10)+log(15)=log(x)

[analog galleon]

log sub 5 125 =3?

[shell condor]

Perfect

[analog galleon]

ah ok

did it wirh no notes

[analog galleon]

understood

[bold zenith]

Try this

[shell condor]

Yup, write the identities too, you'll need them later

[bold zenith]

Do you know about $e$?

[analog galleon]

uhhhh

[soft zealot]

BBMaths

[analog galleon]

isnt it just all those numbers multiplied

[shell condor]

Great!

[bold zenith]

900

[analog galleon]

• turns into x

• turns into division

minus’

[shell condor]

Yup

[shell condor]

Yes

[analog galleon]

I forgot

learned it awhile ago

[bold zenith]

It’s the number used for the natural logarithm ln

That’s the most common base of log

[analog galleon]

ah ok

@shell condor hit me with another

[bold zenith]

But it’s an irrational number, 2.71828…

[shell condor]

Alright, 25^0.5 = 5

[analog galleon]

ahhhh dont confuse me now

[analog galleon]

log base 25 5 =0.5

[shell condor]

That's great!

[analog galleon]

okay

big numbers just switch places

kinda

idk

my brain knows

[bold zenith]

$\ln(720)=a*\ln(2)+b*\ln(3)+c*\ln(5)$

[soft zealot]

BBMaths

[shell condor]

Yup, good one, you gotta remember the power rule here

[bold zenith]

Find a b and c

[analog galleon]

what the hell

[late rose]

prime factorisation?

[shell condor]

$log (a^b) = b*log(a)$

[late rose]

that's what my instinct tells me

[soft zealot]

doctorstrangejr

[bold zenith]

The base is not important here

Yes, try to find the prime factors of 720, hint: they are all 2 3 and 5

[analog galleon]

shit idk

[shell condor]

Yeah, find the powers of 2, 3 and 5 in 720

[analog galleon]

just trying to learn log rn

[shell condor]

Maybe we shouldn't focus on identities yet

[bold zenith]

This might come up tbh

[shell condor]

Let's just stick to the basics

Just write the identities too @analog galleon

[late rose]

are yall helping this guy learn logarithms?

[analog galleon]

I mean the ALEKS gets harder anc harder

cuz it’s an Ai

[shell condor]

Yes

[analog galleon]

goes up to calc they said

I just need a 46 which is college algebra

[late rose]

isnt it just applying the rules?

[bold zenith]

$\ln(2^{a}3^{b}5^{c})=a\ln(2)+b\ln(3)+c*\ln(5)$

[soft zealot]

BBMaths

[late rose]

calculus?

[analog galleon]

whats that

[late rose]

The logarithmic rules

[shell condor]

The power rule, and the others, I'll tag them

[shell condor]

This is the power rule

[analog galleon]

ok

[shell condor]

This is subtraction of logs

[shell condor]

This is addition of logs

[shell condor]

And this is the base identity

[bold zenith]

Ln(1)=0

Log(1)=0

[shell condor]

Write down all of these too, and practise them later

Yup, the only point they intersect at

[bold zenith]

$log_{a}(1)=0$

[soft zealot]

BBMaths

[analog galleon]

this is what I need to get base on my calculator right

[late rose]

Using log instead of lg for log base 10 🥀

[analog galleon]

since I can’t adjust it

[shell condor]

Yes

[bold zenith]

Since timesing by 1 and adding 0 do the same thing it works out

[bold zenith]

Never seen lg

[late rose]

lemme pull up my log notes from some years ago

[bold zenith]

Autocorrect

[shell condor]

We're kinda finished here though, we've all the major identities and the basics, now all there's left is practise

Once you get the basics for log @analog galleon, you just gotta practise the basics and the identities

[analog galleon]

alright

[shell condor]

We've listed the major ones

[analog galleon]

how much time do you have for tutoring left in your day

[shell condor]

Well I should be sleeping 2 hours ago

[bold zenith]

Same

Bye

[shell condor]

Bye mate

[analog galleon]

ah shit im out of luck

EU I presume?

[shell condor]

Nope, Asia

[late rose]

icl there arent many challenging log questions

[shell condor]

I just have a pretty late sleeping schedule

[late rose]

what time is it for you rn?

[analog galleon]

It was just asking me to rearrange it

[late rose]

can you give a sample of a typical difficulty question for logarithms for your exam?

[analog galleon]

well it asked me ig once

it

it was 2^5 =32

the test gets harder or dumber depending how u answer

[late rose]

😭

what

[analog galleon]

it’s the ALEKS

[shell condor]

Sorry dude, can't give the exacts, what I will say is that it is somewhere past 4

[late rose]

oh same

[late rose]

do you have the question?

[analog galleon]

that was the question

2^5 =32 in log

[shell condor]

Well then, I gotta go now

[late rose]

oh

[shell condor]

You're free to send me a friend request @analog galleon if you need help later

[analog galleon]

sleep well

[analog galleon]

yeah, when will you be free?

hours wise

might pay you to be my tutor or something

[shell condor]

Well can't really say, you send me the question and I'll respond whenever I can

[analog galleon]

are you afraid of getting doxxed or something

😭

could just say 6-7 hours

[shell condor]

Dude what🤣, I knew I was fine I never knew I was that good

[late rose]

I mean the concern is fair enough

[analog galleon]

anyone in this server is better than high school math

for the most part

the helpers at least

[analog galleon]

yeah I guess

[late rose]

@analog galleon I got a few challenging questions on logarithms if you wanna try

[analog galleon]

sure

[shell condor]

No I'm being truthful here, I'm going down like a horse dude, stayed awake for 32 hours, then slept for 8, then took a long journey, and now have been awake for around 20 hours

[analog galleon]

give me like 5 mins

[analog galleon]

yooo

We gotta fix that

your math knowledge will start leaking out

[late rose]

If $u = \log_{4}x$, find in terms of $u$.\
(a) $x$\
(b) $\log_{4} 2x$\
(c) $\log_{x} 64$\

[shell condor]

Good ones, remember to manipulate it after converting it into power form for convenience

@analog galleon dude solve these questions, they'll help shape your mind around this concept

[analog galleon]

yeah I will helping my mom

[analog galleon]

X?

cause they switch sides

[late rose]

which part are you doing?

[analog galleon]

Is the answer not x

[late rose]

for part a, the answer is not x.

[late rose]

it's similar to this but in reverse

[final saddle]

@analog galleon Has your question been resolved?

[analog galleon]

I gtg but thx

[final saddle]

Channel closed

Closed by @analog galleon

Use .reopen if this was a mistake.

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[compact trench]

how do you do this?

[compact trench]

i havent done any definite integrals like this that go to infinity yet tho

.-.

[plucky rover]

Substitute u = √y maybe?

[compact trench]

then the u sub would result into a 1/2sqrty 🤔

idk how else to simplify

[plucky rover]

You want to get rid of y lol

What's dy in terms of du

[glossy zephyr]

Ill sadly have to kind of spoil this step, but this one certainly flied over my head and as far as my experience teaching the basics of integration, it does for most people:

$\sqrt{y}=u\\frac{1}{2\sqrt{y}}dy= du$

You probably got here

[soft zealot]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[glossy zephyr]

Theres a reasoning behind it beyond the algebraic fact, but if for some reason your f(y) = u identity appears in the left hand side, you can totally replace it with an u.

[final saddle]

@compact trench Has your question been resolved?

[glossy zephyr]

You could also totally operate with that function to suit whatever your need is, but that usually makes things harder.

[final saddle]

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Available help channel!

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• Show your work, and if possible, explain where you are stuck.
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[radiant carbon]

.close

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Remember:
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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.

[vague anchor]

sorry i meant 16

[frosty rover]

😭 what grade math is that

[vague anchor]

11th

[frosty rover]

IM 11TH ion even understand a single one of those questions 💀

ask mathway

[ivory bear]

cause it is JEE

[frosty rover]

what is JEE

[ivory bear]

internet is a huge domain. help yourself.

[vague anchor]

bro long time no see man

[ivory bear]

@vague anchor crop the question 16 and resend it