My one of my classmates did something even funnier. One time it was r^3 = 216. What he did was turn it into r^3-216=0, and then factored using the difference of two cubes formula, and then took like 5 more lines just to find the quadratic part has no real solution.
What's even better is that to get the answer you need to know what the cuberoot of 216 is, so that you can split it with difference of two cubes, and that's exactly what you're trying to find
The idea is good. I commend your student for that. But if he were to simplify the expression further, he would have to calculate the cubed root of 125 anyways.
I actually want to cry that I can understand this humor, where as a year ago I was totally scared of mathematics and barely passed it, I'm going to finish my business degree, and get another degree in mathematics, thank you.
@@dororj7338 I have seen the US syllabus for maths, it doesn't really make sense to me as most of what you guys study in calc 1 and 2 is Mandatory syllabus in 11th and 12th grade for us, including elementary linear algebra(vector algebra, 3d coordinate system, matrices and determinants), and a decent portion of statistics.
It's funny because I was here thinking "I don't see anything wrong. Isn't this how you are supposed to solve this equation?" Then I looked at the comments, realized the joke and then realized the simplicity of the question.
Lmfao i was literally thinking "hm? what's the problem here?", don't think i would have done it this way, but i didn't realize that it was overcomplicated for some reason
Well actually a cubic equation has three equations, only 2 of them are complex roots (so solutions that have imaginary numbers), and real root being 5.
I wish I understood this quick trick back in undergrad in the late 2000's when I was taking calculus courses. Still passed but it was literal torture. My math professor wasn't very good at explaining. Basically, the steps when an exponent on a variable is involved is to: - take _ln_ on both sides so exponent can be brought down - isolate the variable - take _e_ on both sides to cancel out the _ln_ on the variable - now just do basic algebra to solve for the variable
@@anshumanagrawal346 The logarithm is a multivalued function, but we only take the principal branch. It's the same thing how the square root is multivalued, but we only take one value when calculating √x --- the principal square root
@@anshumanagrawal346 ln(125) = ln(125e^(2inπ)) = Ln(125) + 2inπ, n ∈ ℤ ln(125)/3 = Ln(125)/3 + 2inπ/3 e^(ln(125)/3) = e^(Ln(125)/3)*e^(2inπ/3) = (³√125)e^(2inπ/3) n = 0 gives the real value of ³√125, n = 1 gives one imaginary root, and n = 2 gives us the other. All other values of n are just repeats of the roots we've already found
Because: log a^2 is just log a + log a which is also 2 log a log a^3 is just log a + log a + log a which is 3 log a log a^4 is just log a + log a + log a + log a which is 4 log a The exponent will be the amount of times u add the logarithms together, which is the same as multiplying the exponent by the logarithm. This isn’t a proof or anything but hopefully it explains it better
That was a good method, learned alot from this little thing. But not sure why people want to simplify everything; sometime doing the hard way gives alot of experience.
You could also simplify 125 to 5^3 and put the three in front of the ln so that you can simply the 3 below. Edit: or you could just avoid using the ln, simplify 125 to 5^3 and put both members under the cubed root so the ^3 goes away.
Appreciate the creativity, but if you're going to use a calculator anyway at some point of the process, might as well use it at the beginning for cube root.
x³=125 I pose x=a+bi (with a and b two reals numbers) (a+bi)³=125 a³+3a²bi-3ab²-b³i=125 (a³-3ab²)+(3a²bi-b³i)=125+0i So we have the system: a³-3ab²=125 3a²b-b³=0 b(3a²-b²)=0 So either b=0 or 3a²-b²=0 if b=0 we have a³-3ab²=a³=125 (with a a real number) a1=5 and b1=0 x1=a1+b1×i=5+0i=5 If b≠0 3a²-b²=0 b²=3a² So a³-3ab²=a³-3a²×3a²=-8a³=125 a³=-125/8 a= -(5/2) a²=25/4 3a²=75/4 b²=3a²=75/4 so b=sqrt(75)/2 or b=-sqrt(75)/2 So we have a1=5. b1=0 a2=-5/2. b2=sqrt(75)/2 a3=-5/2. b3=-sqrt(75)/2 x1=a1+ib1=5+0i=5 x2=a2+ib2=(-5+i×sqrt(75))/2 x3=a3+ib3=(-5-i×sqrt(75))/2
If we are working with real numbers there's only 1 solutions, but there are 2 More in complex, Let me get all answers. x³=125 ⬜Substract 125 on both sides x³-125=125-125 x³-125=0 ⬜Rewrite 125 as 5³ x³-5³=0 ⬜A difference of cubes a³-b³ can be written as (a-b)(a²+ab+b²) Letting a=x, b=5 x³-5³=0 (x-5)(x²+x•5+5²)=0 (x-5)(x²+5x+25)=0 ⬜If ab=0, a=0 or b=0 x-5=0 or x²+5x+25=0 ◼️Solve for x-5=0 x-5=0 ⬜Add 5 on both sides x-5+5=0+5 x=5 ◼️Solve for x²+5x+25=0 by formula Formula: x=[ -b±√(b²-4ac) ] / 2a Having a=1, b=5, c=25 x²+5x+25=0 x =[ -5+√(5²-4(1)(25)) ] / 2(1) x =[ -5±√(25-4(25)) ] / 2 ⬜Factor 25 x =[ -5±√(25(1-4)) ] / 2 x =[ -5±√(25•(-3)) ] / 2 ⬜Rewrite 25 as 5² and expand the root x =[ -5±√5² • √-3 ] / 2 ⬜ Rewrite -3 as 3•(-1) and expand the root x =[ -5±5√3 • √-1] / 2 ⬜i²=-1 or i=√-1 x =[ -5±5√3 • i] / 2 x =[ -5±5i√3 ] / 2 Therefore x₁ = 5 x₂ = (-5+5i√3) / 2 x₃ = (-5-5i√3) / 2 1 real solution 2 complex solutions, i² = -1
Yes this would be the first question in the most basic non-calculator exam. If you do higher, and see something like this, you're done for (foundation kids will get better so it's harder for the higher kids to get even better). Lol
"And now we can just use our calculator"
You don't really need a calculator. Rewrite 125 as 5^3. Bring the 3 down and cancel with the 3 in the denominator. Then you have e^ln(5) which is 5.
@@onyeka2265 if you rewrite 125 as 5^3 you don't even need logarithms lol
@@flavius9899 noo, non l'avrei mai detto
@@flavius9899 you could also rewrite it at 125 ^ 1/3 (or the cube root of 125), ln cancels out, and you get 5!
@flavio Sorbo you missed the joke *this* hard
when you go back to the first level after beating the game:
Lol
Lol
XD
ahh this is so accurate
After playing celeste, this is SO ACCURATE
That's how your brain functions after studying college level maths, can't help getting a little fancy
can relate
this is what i literally do whenever i see simple problems on the internet lol
@@stefanchandra1237 nice username. I'm top fan of Your Reality track on Spotify with over 600 times played last year
True but this is not college level math?
@@zurps high school
Cocky freshman yeah
When you become a software engineer and want to print hello world
mov rdi, 1
mov rsi, "Hello world!"
mov rdx, 12
mov rax, 1
syscall
Public class main { public static void main(String[] args) { System.out.println("Hello, world");}}
#include
int main(int argc, char** argv) {
printf("hello world!");
return 0;
}
Amateurs. Open notepad. Type "Hello World".
Press ctrl + p and press enter!
#Include
Using namespace std;
int main(){
Cout
My one of my classmates did something even funnier. One time it was r^3 = 216. What he did was turn it into r^3-216=0, and then factored using the difference of two cubes formula, and then took like 5 more lines just to find the quadratic part has no real solution.
i mean... he could get the imaginary solutions that way, I guess?
he has a video that did that with x^3 = 1 lol. forgot if it was on fast! or bprp
What's even better is that to get the answer you need to know what the cuberoot of 216 is, so that you can split it with difference of two cubes, and that's exactly what you're trying to find
@@sartajdhaliwal9462 one of them should be the real Answer of that
Nah we prove and use cubic formula
The idea is good. I commend your student for that. But if he were to simplify the expression further, he would have to calculate the cubed root of 125 anyways.
I think that just might be the point of this video
Everyone knows 5 cubed is 125
So its just waste of time ig
Or they did it for fun
@@jeeteshsawarkar1808 r/woosh 😄
I think it's just how they have to prove it. You can't just write the answer without showing how you got it. Maybe I'm wrong.
X^3=125
X^(3) = (1+2)5
X=5
P/s: I didn't think I'd get this far...
Wtf? So (1+2)*5 =125?
@@justarandomlol exactly
What
No… you would just take the cubed rot if 125 which is 5 so x=5
Habibi method 😂
Alt title: how real man solve functions
That's another channel
@@volodymyrgandzhuk361 yeah, it's F********** M******
@@ghijklabcdefflammable maths?
Khaby then comes in with the cube root and solves it in one step
2 ROOT PIE OVER V OF X WHERE IS PIE> WHAT I DONT UNDERSTAND IS HOW MANY LETTERS ARE IN YOUR ALPHET?? DID I SPELL THAT RONG... LOL
@@robbyandrews6318 sheesh u gotta chill m8
Continued
x=cbrt(e^ln(125))
x=cbrt(125)
x=5
Does cbrt means cube root?
@@Nano-n no, it means Cuberto, the digital design and development agency.
@@boston5814 ah damn, it all makes sense now! thanks a lot
@@abhimanyupandey8170 yep no problem, any more questions, send them this way!
@@Nano-n ofc
I actually want to cry that I can understand this humor, where as a year ago I was totally scared of mathematics and barely passed it, I'm going to finish my business degree, and get another degree in mathematics, thank you.
I mean... It's just highschool level maths though. Every 16-17 year old knows this
@@dororj7338 I have seen the US syllabus for maths, it doesn't really make sense to me as most of what you guys study in calc 1 and 2 is Mandatory syllabus in 11th and 12th grade for us, including elementary linear algebra(vector algebra, 3d coordinate system, matrices and determinants), and a decent portion of statistics.
@@TheMartian11 well they don't teach logs or calculus in the UK, so to understand this humor in a couple of months it's not bad.
@@liviu445 They do; the whole of A-Level is pretty much based on calculus.
@@williamallen9145 Yeah, i was dumb while i made this comment.
The roots of the equation are 5, 5ω, 5ω^2 (where ω= (-1±√3i)/2
when they have access to calculator and dont use cube root
It's funny because I was here thinking "I don't see anything wrong. Isn't this how you are supposed to solve this equation?" Then I looked at the comments, realized the joke and then realized the simplicity of the question.
smh here's the full proof x= e^(ln(125))/3
Therefore x =e^((ln(125))1/3
x =125^1/3
this is obviously 5 QED
I love how you cut that video in the end
"wtf is a cube root" -bprp probably
oh thnx now i can use the calculator ...initially it was hard
Lmfao i was literally thinking "hm? what's the problem here?", don't think i would have done it this way, but i didn't realize that it was overcomplicated for some reason
It could also be like
Ln x³ = ln 125
3 ln x = ln 5³
3 ln x = 3 ln 5
Ln x = ln 5
By comparing x = 5
"Life is Simple , we make it difficult "
😂😂😂😂
Or just cube root
Me: "5, no need for unnecessary equations)
The teacher: Only my method will be followed in the exam
The teacher's method
Dude, you missed two complex roots.
Seems to be a solution geared toward first-year algebra students.
This is the easiest short cut trick I have ever seen in my life for finding cube of a number
Meanwhile
Cube Root: 🙄
Well actually a cubic equation has three equations, only 2 of them are complex roots (so solutions that have imaginary numbers), and real root being 5.
Me in the 8th grade who just cube rooted it and found the answer without calculator- I like your funny word magic man
"smart people don't understand jokes"
-sun tzu the art of mathematics
Anime nerd
@@molk1217 thank you
@@kechan_ thank you denied
well, I guess I'm something of a genius myself
Me too
me: memorized cubes till 10
From now onwards I will not able to convince myself directly that solution of x³=125 is 5
The most efficient solution
This is a very useful beginner introduction to logarithms!
bro i really just accidentally found the answer by replacing x to 5
I love how u made it like 5 times more complicated than it should be.
5³ more complicated?
Wow, 60seconds of my life that I'll never get back
+ the time you spended writting this comment and maybe reading this one
Do you usually get time back? What makes these 60s remarkable?
Future Ultron after watching this video :
Self-conscious lvl 100
Just remember to use logs only when the exponent is a variable, that usually works for me
First thought: Yeah, it's 5 and now?
This has 3 solutions.
wow its an interesting discovery that e to the 3 times logx is just 3rd root of x
Logarithm properties, logx^k = k*logx and e^logx = x
Funny how I was just taught natural logarithm 2 hours ago and now this is on my recommendations
I just graduated elementary, but I'm gonna save this video for when I understand what on earth he's saying
We could also just find the cube root lol
This stresses me out to the maxxxx!!! Where did the ln come from?! 😭
This is exact moment blackpenredpen became student.
When you become a head chef and want to make instant ramen :-
Lmaooo
Wait which step loses complex answers?
e^ln(125)/3 = 5
The audio cut at the end is *immaculate* 💀
Now i know what my dog feels when i talk to him
0:18
殺雞偏用牛刀 XD
重點是最後還用calculator 整個笑死
Cant we just transpose that cube to other side to give a cube root of 125?
I wish I understood this quick trick back in undergrad in the late 2000's when I was taking calculus courses.
Still passed but it was literal torture. My math professor wasn't very good at explaining.
Basically, the steps when an exponent on a variable is involved is to:
- take _ln_ on both sides so exponent can be brought down
- isolate the variable
- take _e_ on both sides to cancel out the _ln_ on the variable
- now just do basic algebra to solve for the variable
How come we lose the complex solutions by doing this?
Implicitly using the principal branch of the log selects a solution.
@@materiasacra oh, so you mean that we can't write ln(x^3) as 3ln(x) always? Or that we can't "cancel" it with e?
@@anshumanagrawal346 The logarithm is a multivalued function, but we only take the principal branch. It's the same thing how the square root is multivalued, but we only take one value when calculating √x --- the principal square root
@@gamerdio2503 I know that, but my question is which step exactly assumes the principal branch
@@anshumanagrawal346 ln(125) = ln(125e^(2inπ)) = Ln(125) + 2inπ, n ∈ ℤ
ln(125)/3 = Ln(125)/3 + 2inπ/3
e^(ln(125)/3) = e^(Ln(125)/3)*e^(2inπ/3) = (³√125)e^(2inπ/3)
n = 0 gives the real value of ³√125, n = 1 gives one imaginary root, and n = 2 gives us the other. All other values of n are just repeats of the roots we've already found
Can someone explain why u can move the exponent to the front when there is a ln
It's a log rule
log a^b = b log a
yeah its a log rule, i recommend you to proof it by yourself,or find the proof online, i've watched sal khan's video on the proof (khan academy)
e^(ln(a^b))=a^b,
a=e^(ln(a^b)/b),
ln(a)=ln(a^b)/b,
bln(a)=ln(a^b)
Because:
log a^2 is just log a + log a which is also 2 log a
log a^3 is just log a + log a + log a which is 3 log a
log a^4 is just log a + log a + log a + log a which is 4 log a
The exponent will be the amount of times u add the logarithms together, which is the same as multiplying the exponent by the logarithm.
This isn’t a proof or anything but hopefully it explains it better
"WHAT THE F#CK IS A CUBE ROOT 🦅🦅🦅"
This is how i am solving my life problems😅
Love this format
Cube root of 125 is 5
So 5 cube is 125
Solved
That cutoff was gold
I’d love to see this with an example I wasn’t able to pick up on at the same sped at which I know my times tables
When you are getting bored in class
dude's making it more complicated than it actually is.
High school kid be like - that's 5
...he starts explaining....
Kid - wtf!!
every high school kid knows log....
spent forever looking for this video
Congratulations you made the question much harder than it actually is.
" I've seen this one it's a classic "
*avg Asian enters*
I need to really relearn algebra first and then go back into deeper knowledge
You can just use the cube root method right? Or mental maths for small quests like that so 5.
That was a good method, learned alot from this little thing. But not sure why people want to simplify everything; sometime doing the hard way gives alot of experience.
The true asians have already memorized these by heart😂
ITS JUST THE CUBE ROOT OF 125 OMG
why did i feel like this would be more efficient than a cube root for a solid 10 seconds....
This method can be transferred to many other problems. That is why he is showing it like that.
the way it gets cut off as he says "five" makes this even funnier
I'm happy, finally I understood something on this channel
Cool but I’m thinking how about getting two other roots (5*(-1+-sq(3)i)/2) by logarithm?
When you are too advanced in math and eventually forgot roots be like:
The design is very human
Making Yout Life Harder With Math Be Like :
When you just happened to know that 5^3=125
Guys it’s a joke stop saying he’s doing it wrong just admire the creativity.
You could also simplify 125 to 5^3 and put the three in front of the ln so that you can simply the 3 below.
Edit: or you could just avoid using the ln, simplify 125 to 5^3 and put both members under the cubed root so the ^3 goes away.
Appreciate the creativity, but if you're going to use a calculator anyway at some point of the process, might as well use it at the beginning for cube root.
Bro escaped the Matrix
x³=125
I pose x=a+bi (with a and b two reals numbers)
(a+bi)³=125
a³+3a²bi-3ab²-b³i=125
(a³-3ab²)+(3a²bi-b³i)=125+0i
So we have the system:
a³-3ab²=125
3a²b-b³=0
b(3a²-b²)=0
So either b=0 or 3a²-b²=0
if b=0 we have
a³-3ab²=a³=125 (with a a real number)
a1=5 and b1=0
x1=a1+b1×i=5+0i=5
If b≠0
3a²-b²=0
b²=3a²
So a³-3ab²=a³-3a²×3a²=-8a³=125
a³=-125/8
a= -(5/2)
a²=25/4
3a²=75/4
b²=3a²=75/4
so b=sqrt(75)/2 or b=-sqrt(75)/2
So we have
a1=5. b1=0
a2=-5/2. b2=sqrt(75)/2
a3=-5/2. b3=-sqrt(75)/2
x1=a1+ib1=5+0i=5
x2=a2+ib2=(-5+i×sqrt(75))/2
x3=a3+ib3=(-5-i×sqrt(75))/2
If we are working with real numbers there's only 1 solutions, but there are 2 More in complex, Let me get all answers.
x³=125
⬜Substract 125 on both sides
x³-125=125-125
x³-125=0
⬜Rewrite 125 as 5³
x³-5³=0
⬜A difference of cubes a³-b³ can be written as (a-b)(a²+ab+b²)
Letting a=x, b=5
x³-5³=0
(x-5)(x²+x•5+5²)=0
(x-5)(x²+5x+25)=0
⬜If ab=0, a=0 or b=0
x-5=0
or
x²+5x+25=0
◼️Solve for x-5=0
x-5=0
⬜Add 5 on both sides
x-5+5=0+5
x=5
◼️Solve for x²+5x+25=0 by formula
Formula: x=[ -b±√(b²-4ac) ] / 2a
Having a=1, b=5, c=25
x²+5x+25=0
x =[ -5+√(5²-4(1)(25)) ] / 2(1)
x =[ -5±√(25-4(25)) ] / 2
⬜Factor 25
x =[ -5±√(25(1-4)) ] / 2
x =[ -5±√(25•(-3)) ] / 2
⬜Rewrite 25 as 5² and expand the root
x =[ -5±√5² • √-3 ] / 2
⬜ Rewrite -3 as 3•(-1) and expand the root
x =[ -5±5√3 • √-1] / 2
⬜i²=-1 or i=√-1
x =[ -5±5√3 • i] / 2
x =[ -5±5i√3 ] / 2
Therefore
x₁ = 5
x₂ = (-5+5i√3) / 2
x₃ = (-5-5i√3) / 2
1 real solution
2 complex solutions, i² = -1
I thought this was a without a calculator question 😭😭😭
Yes this would be the first question in the most basic non-calculator exam. If you do higher, and see something like this, you're done for (foundation kids will get better so it's harder for the higher kids to get even better).
Lol
Me knowing that it is 5 by looking at it:
"Sometimes my genius is almost frightening"
I watched this in slow mo and still couldn't keep up!😂
Why not using the cubic root ?
I dig Great in Algebra, and fell off the planet when I went to logarithms , but that was 40 years ago
Wouldn't it be easier to use the calculator to get the cube root of 125?
If you consider that ln(125) = 3ln(5) the 3s cancel and the ln cancels with the e to give the answer 5.
This is the best and only method for this. What’s a cube root?
When you know it’s 5 right away and feel like a galaxy brain
Driving from Philadelphia, PA to New York city through Los Angeles... there 2 more roots... total 3 roots...
You can just keep dividing by five too since we know that 125 is a perfect square and 5 is the unit's place.
Cubic Root is crying on the edge of the world
Even switching guns is slower than you switching pens
At least you could have found also the 2 other complex solutions