Those who follow channels like blackpenredpen, brithemathguy and prime newton's already know such functions. They use things like Dirchlet Eta function, Gamma function, Beta function etc. extensively in their problems.
I started by noticing that xe^x is inside the function and in the equation with K we have a ln(2) so if we put some ln(2^y) in the lambert function we can get in the form of the K equation so let k = ln(2^y) so, ln(2^y) x e^{ln(2^y)} = 32(ln2) yln(2) x 2^y = 32(ln(2)) y(2^y) = 32 then taking log 2 on both sides log2(y) + y = 5 this is somewhat similar to the first eq so if we substitute y = 2^x we get the first eq again hence y = 2^x k = ln(2^{2^x}) k = (2^x)(ln(2) since 2^x = 5-x we can also write as k = (5-x)(ln(2))
My approach was to use k.e^k = 32ln2. So I took k = m(ln2) (say) Then we get m(ln2).e^(m(ln2))=32ln2 m.2^m=2^5 Taking log to the base 2 on both sides, we see that x = log m to the base 2 So x + m = 5 and m = k/ln2.
Mujhe pehle kuch nahi samajh raha tha ki kaise approach karu, then maine aapka hint suna aur phir maine x ko log base 2 (2^x) likha and 5 ko log base 2 (2^5) likha then dono ko same side pe leke gaya and used the difference of log property uske baad log2(a) = ln(a)/ln(2) wali property use kari to equation bana: ln2.2^x = ln(2^5/2^x) and phir mujhe dikha ki agar mai 2^x LHS se RHS me likhdu aur dono side 32 se multiply kardu to exact u.e^u wala form bana aur K bhi ban gaya jisme u = ln(2^5/2^x) and phir mujhe same answer mil gaya 😁
I have a different approach I saw 32 and it screamed out 2 to the power 5 to me So I wrote this 2^(x+2^x)=2^5 (2^x)e^((2^x)ln2)=32 Now multiplying by ln2 and taking Lambert w of both expressions we get 2^x=k/ln2 ab initial exppresion se 2^x ko 5-x likho and ans paao Basically I took the whole exppresion and kept it on the power of 2 which gave the ans
From the first equation of Lambert function Substituting x--> (ln x) We get W(lnx.x)=lnx Now...writing R.H.S and L.H.S as powers of 2.. 2^(x) . 2^(2^x) = 32 Multiplying ln2 both side 2^(x)ln2 . 2^(2^x) = 32 ln2 Taking Lambert function both side.. W(ln 2^(2^x) . 2^(2^x) ) = k [A.T.Q] We can write left side as ln (2^(2^x)) = k Simplifying...we get x= log²k - log²lnk (Log² matlab log base 2... Phone me subscript pe kaise likhte hain pata nehi 😅...btw is it correct? )
Lambert W function ki jrurat Jee me nhinhogi kbhi bhi I guess kyuki uske liye Lambert W ki values yaad karni padti h, lgta ha bhaiya ne bas khi yt pe dekh liya aur bta diya lmao
bhai abhi tu bas apni polynomial ke jo manipulation type questions hote hai unko master karle. fir seedha 11th le lecture dekhna proper order mein. koi random/ easy chapter lene se fayda nahi hoga
@@IcY200 mat kar bhai. properly 11th padhlena usse jaada fayda hoga. calculus 12th mein aata hai toh 11th ki maths tumlogo ko samajh aayegi. detail mein padh shuru se
explain them the need, they’ll understand. This year so many students are getting above 200+ in so60 community, I’ll be amazing to have u as a part of it 😄
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Those who follow channels like blackpenredpen, brithemathguy and prime newton's already know such functions. They use things like Dirchlet Eta function, Gamma function, Beta function etc. extensively in their problems.
Yeah
Yess
Yessss :) bprp is like insanee ❤
I'm happy that someone finally addressed Prime Newtons, he is quite underrated
Forgot 3b1b
Haha lol many ytubers have made video on this one. You're the first one coming up with this on jee related channel.
Surely a W move
Quite literally
Damnnn I'm so proud I already knew this function for months before it's good to see you making video on this
I started by noticing that xe^x is inside the function and in the equation with K we have a ln(2) so if we put some ln(2^y) in the lambert function we can get in the form of the K equation so let k = ln(2^y)
so, ln(2^y) x e^{ln(2^y)} = 32(ln2)
yln(2) x 2^y = 32(ln(2))
y(2^y) = 32 then taking log 2 on both sides
log2(y) + y = 5 this is somewhat similar to the first eq so if we substitute y = 2^x we get the first eq again
hence y = 2^x
k = ln(2^{2^x})
k = (2^x)(ln(2)
since 2^x = 5-x we can also write as
k = (5-x)(ln(2))
My approach was to use k.e^k = 32ln2.
So I took k = m(ln2) (say)
Then we get
m(ln2).e^(m(ln2))=32ln2
m.2^m=2^5
Taking log to the base 2 on both sides, we see that
x = log m to the base 2
So x + m = 5 and m = k/ln2.
me who watched bprp videos and "i knew"...but solve nhi hua..😁
Graphical method
1.Intersection of straight line
y=5-x
With
2. Exponential function graph 2^x
bina GDC ke?
bhaiya ye thumbnail thoda cringe ni h?
Darawana jyada h
The kid in the thumbnail is a so60 member
@@somil3368 isse badhiya kratos ko thumbnail mein rakh dete thoda relatable sa lagta
Poora video thoda cringe nahi hai?
@@brijeshsrivastava6682 ha bhai marks kitne aree woh bta
Aa jate koi bhi
Bhaiya my answer is same but little but lengthy I solve in different way but yeh good question 😊
Mujhe pehle kuch nahi samajh raha tha ki kaise approach karu, then maine aapka hint suna aur phir maine x ko log base 2 (2^x) likha and 5 ko log base 2 (2^5) likha then dono ko same side pe leke gaya and used the difference of log property uske baad log2(a) = ln(a)/ln(2) wali property use kari to equation bana: ln2.2^x = ln(2^5/2^x) and phir mujhe dikha ki agar mai 2^x LHS se RHS me likhdu aur dono side 32 se multiply kardu to exact u.e^u wala form bana aur K bhi ban gaya jisme u = ln(2^5/2^x) and phir mujhe same answer mil gaya 😁
I have a different approach
I saw 32 and it screamed out 2 to the power 5 to me
So I wrote this
2^(x+2^x)=2^5
(2^x)e^((2^x)ln2)=32
Now multiplying by ln2 and taking Lambert w of both expressions we get
2^x=k/ln2 ab initial exppresion se 2^x ko 5-x likho and ans paao
Basically I took the whole exppresion and kept it on the power of 2 which gave the ans
K can be found by replacing x by lnx in the lambert eqn then putting x=8
Ab toh lambert function ki bhi practice karni padegi🥲
Koi nahi manipulation improve karne ke liye accha lag raha hai
Jee apirants truely seeking for advanced level know this channel
ye starting me ladki ka avaj to nahi lag raha
0:10 bhaiya aapki aawaj ko kya ho gya??
Senda namak
my approach: mene pehle step me he x ko 5 - t se substitute kr diya tha fir apna aap he ye is langur function ki form me aa gya tha
I used that given valued function for k
W(33.ln2 ) ko factorise karke x.e^x banana ki koshish ki
There were 3-4 questions in some allen assignment on lambert w function
Bhiaya helium phok li kya ??
From the first equation of Lambert function
Substituting x--> (ln x)
We get W(lnx.x)=lnx
Now...writing R.H.S and L.H.S as powers of 2..
2^(x) . 2^(2^x) = 32
Multiplying ln2 both side
2^(x)ln2 . 2^(2^x) = 32 ln2
Taking Lambert function both side..
W(ln 2^(2^x) . 2^(2^x) ) = k [A.T.Q]
We can write left side as ln (2^(2^x)) = k
Simplifying...we get x= log²k - log²lnk
(Log² matlab log base 2... Phone me subscript pe kaise likhte hain pata nehi 😅...btw is it correct? )
Maine same method use kiya
Shouldn't it be (ln2). 2^x . 2^(2^x) inside Lambert w
@NirbhayJEE2025 yes..! m lnx = ln (x^m) property lagana padega form me aane ke liye
Kyuki isse W( ln t . t ) form me laana hain
@@Vabadrish but fir Mera expression aur apka expression thoda alag aa rha hai
its fascinating that i knew about this already huehue
Mast question
Bruh, why is lambert function coming in jee???!!!!1
Ni bhai video dekho poori😂
Views badhani ki ninja technique
Problem solving session kaha gaya??
mujhe pata tha yeh blue pen red pen par bhout dekha hai
Juat harvard problem dekhke aaya hu Prime Newtons se, So cant watch this again
Ye toh prime newton ke channel pe dikha hai lol
Bhaiaya if we solve like this
Given x+2^x =32
taking both term in power of two
2^x × (2^x)^2^x=32
Then putting 2^x=V
I AM getting different soln for x
Prime Newton ☠️☠️☠️
I saw it on bprp
haha blackpenredpen ka video dekha hua tha lambert function pe (islie easily solve hogaya)
Whi, easy tha
Ye kis jee mains ke paper mei aagaya?
Lambert W function ki jrurat Jee me nhinhogi kbhi bhi I guess kyuki uske liye Lambert W ki values yaad karni padti h, lgta ha bhaiya ne bas khi yt pe dekh liya aur bta diya lmao
a student shared this, and i can create any function with such property as the logic strongly lies in JEE Domain
Bhaiya mai 10 me hu 11 iss session me jaunga kon kon se aise chapters jo mujhe complete karni hogi fast syllabus completion. Mtlb hardest one
Mai bhi 10th mai Hu mera advice hoga asaan chapters se shuru karo aur mastery karo jese sets
Rn I’m doing pnc
bhai abhi tu bas apni polynomial ke jo manipulation type questions hote hai unko master karle. fir seedha 11th le lecture dekhna proper order mein. koi random/ easy chapter lene se fayda nahi hoga
@@IcY200 mat kar bhai. properly 11th padhlena usse jaada fayda hoga. calculus 12th mein aata hai toh 11th ki maths tumlogo ko samajh aayegi. detail mein padh shuru se
Do P&C properly
Solved 😀
not to brag bout it but i know it as a 10thie and its cool. gamma function and all that stuff
Bprp supremacy
I knew 😏
did the exact same thing 4:41 fir aagey nahi badh paya
Ye to aata h mere ko
W(🐟 e^🐟) = 🐟
I am Avg bprp viewer
😂
Bhaiya mai so60 ke que Krna toh chahta hu but itne paison ke lia parents nhi mane ge 😢
But it's ok at the end apko bhi source of income chaiye hi 😊
explain them the need, they’ll understand.
This year so many students are getting above 200+ in so60 community, I’ll be amazing to have u as a part of it 😄
@jeesimplified-subject ?
@@jeesimplified-subject bhaiya can i get a set of good questions ?
Bprp viewers🗿🗿
True
true😅
Ans; X=5-k otherwise aap complex se kroge to h infinitely many soln aa jayenge
Lambert W function
W function 😂
esea cya
this is reverse clickbait: great video but awful thumbnail 🙏
i’d admit 😂
First