Solve Yourself, Before watching the video

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Solve Yourself, Before watching the video

Рет қаралды 60,567

Ай бұрын

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@jimmywatson7950 Ай бұрын

Sir, In the equation 4ˣ=x⁶⁴, both 4ˣ and x⁶⁴ are continuous function if x=1, then 4¹>1⁶⁴ if x=2, then 4²

@sandippatwari927 17 күн бұрын

By bisection method

@ABCDEFGH4567-8 Ай бұрын

Set f(x)=4ˣ-x⁶⁴ Clearly x=256 is solution. f(0)=1>0 f(1)=3>0 f(2)

@Qwert-c1 Ай бұрын

😂

@kx_lovestatus7175 Ай бұрын

😂😂 baag aya se

@arkodasgupta0412 Ай бұрын

@@kx_lovestatus7175 why laughing bro? Jo roots non solvable hote ha manually, computers use this algorithm for root finding. This is a very basic one. There are other ones such as newton raphson which converge much faster

@SAZAEMAUTPW Ай бұрын

@@arkodasgupta0412 bhaiuya newton rapshon method kab padhaya jata hai i mean iit ke syll. me hai ??

@ABCDEFGH4567-8 Ай бұрын

@@arkodasgupta0412 pata nahi hoga usko... isliye has raha hai.... Numerical analysis/ methods naam ka koi branch hai mathematics ka sayad nahi pata hoga..may be that's why he is laughing

@AbhinabRay-tf5gh Ай бұрын

Vaah sir kya logic hain. Love you very much sir❤❤❤❤

@rahulmahato4072 11 күн бұрын

4^x =x⁶⁴ xlog4=64logx 1/64 log4=1/x logx 4/256 log4=1/x logx 1/256 log4⁴ =1/xlogx x=256😊

@sangeeta3729 20 сағат бұрын

good one

@JoaoPedro-js6mk Ай бұрын

Hame to sir ne ye Kara dia tha wo sare coaching teachers ke association me h maths ke islie unhe daravne question milte rehte h or hme dete h taki hmari aukat pata chale...

@SantoshChoudhary-vz5yg Ай бұрын

We could use log here and Equation that we get is By taking log base 2 on both side x=32log(x) x=16*logx² We can now analyse the equation and get that X must be greater than 1 Try to make the equation symmetrical by putting logx²=16 Hence we get x=2⁸ We can now varify our solution 2⁸=2⁴x2⁴ Hence x=256

@NamanAggarwal87363 Ай бұрын

2nd solution 4*x = 4* 64log x at base 4 ||. x =64 log x at base 4 x/64 = logx at base 4 Now by hit and trial x =256 satisfies

@Takhatsinghcharan-ge2wn 12 күн бұрын

Taking log both side to the base 2 ,we get X=32 log X base 2 Now log X base 2= k So X = 2^k Put 2^k= 32 k 2^(k-5)=k Clearly k= 8 So X=2^k=2^8=256

@taseen5945 Ай бұрын

sir , i was the type of student to think to bring log in first attemp, such a beautiful way to solve

@aarjav4138 Ай бұрын

I Actually Solved It Using Basic Log Only🤷‍♂️

@anirudhsingh1623 Ай бұрын

Can you share the solution plz 🙏🏻

@zeroplays9915 Ай бұрын

@@aarjav4138 then what youll have to use Lambert W function which is not in jee syllabus

@uncannyroaches5933 Ай бұрын

xlog4= 64 logx logx/x= (log 4)/64 logx/x= (4log4)/256 logx/x= log 256/256. easy peasy x= 256

@dheeraj76 Ай бұрын

Sir please upload old videos of class 11 and 12.😢😢😢😢😢

@ABCDEFGH4567-8 Ай бұрын

4ˣ=x⁶⁴ Taking both side log xln4=64lnx ln4/64=lnx/x lnx/x=ln2/32 (1/x)ln(1/x)=(-1/32)ln2 Taking both side W-lambart function. W[(1/x)ln(1/x)]=W[(-1/32)ln2] ln(1/x)=W((-1/32)ln2) 1/x=e⁻ʷ⁽⁽⁻¹/³²⁾ˡⁿ²⁾ x=eʷ⁽⁽¹/³²⁾ˡⁿ²⁾ x=1.02 x=-0.97 x=256

@MaheshKumar-lx1ku Ай бұрын

Ye function Kab padhaya jata hai??

@snowball1002 Ай бұрын

@@MaheshKumar-lx1ku not in jee syllabus

@shivx3295 Ай бұрын

Lambard W function ke baare mai kaha se seekha?

@tmrapper6378 Ай бұрын

@@shivx3295professor Newton's

@MaheshKumar-lx1ku Ай бұрын

@@snowball1002 ohhh...

@ipshitpaul Ай бұрын

X=32log2(x) X=256 satisfies this equation... Let's assume f(x) =x-32log2(x) F`(x) =1-32/x.log2... This is clearly decreasing in the interval (256, infinity)

@zeroplays9915 Ай бұрын

how did you guess x=256 either youre a genius or a cheater then

@ManishKoiri-qw3xm Ай бұрын

By taking log both sides directly

@zeroplays9915 Ай бұрын

then what youll have to use Lambert W function which is not in jee syllabus

@rajshauryz Ай бұрын

Maine log leke hit and trial se kardiya (X)^(32/x)=2

@veydantliit8952 Ай бұрын

Bhai tere auto correct ne teri sacchai dikha di😂

@Idk2007-fx4cs Ай бұрын

@@veydantliit8952pehele kya tha

@zeroplays9915 Ай бұрын

then what you'll have to use Lambert W function which is not in jee syllabus

@harshpandey8528 Ай бұрын

Not defined because a≠0 a>0

@SamyakMeshram1 Ай бұрын

Sir, please 😟 make a video on the question : Q. The equation (x ^ 2 + x + 1) ^ 2 + 1 = (x ^ 2 + x + 1)(x ^ 2 - x - 5) for x \in (- 2, 3) will have number of solutions, (1) 1. (2) 2. (3) 3. (4)Zero. Sir I waiting for your video. 🙂🙂🙂🙂🙂

@vishalachakravarthy2082 9 күн бұрын

Very much impressed sir . thank you for this knowledge sir lots of respect to you sir

@PiyushKumaryadav-yd5ft Ай бұрын

4^x =x⁶⁴ 4 = log base(x) arg (x⁶⁴) Now 4=64log(base x) arg(x) Then 4 = 64 😮😮😮😮😮 So equation is wrong because it is not possible

@charushashi 7 күн бұрын

Initially comparing the function with graph there will be two solutions

@litonbiswas9018 Ай бұрын

4^x=x^64 =>xln4=64lnx =>lnx/x=ln4/64 =>lnx/e^lnx=ln4/64 =>-lnx.e^-lnx=-ln4/64 =>W(-lnx.e^-lnx)=W(-ln4/64) [taking lambert W function in both sides] =>-lnx=W(-ln4/64) =>lnx=-W(-ln4/64) x=1/e^W(-ln4/64) ans.. This qsn can be solved in such method...

@kushagramishra1729 Ай бұрын

lekin sir 256 ke alawa 2 aur solutions hai iske ek -1 se 0 ke beech me aur ek 1 se 2 ke beech me

@aryankushwaha7028 Ай бұрын

Yes i also think so that's why I analyze behaviour , after I plot the graph on desmos then i got to the point

@kushagramishra1729 Ай бұрын

@@aryankushwaha7028 There are totally 3 real solutions. One of them is 256

@ABCDEFGH4567-8 Ай бұрын

@@kushagramishra1729 Use the numerical analysis to find the other 2 solutions. Apply the bisection method

@Aadi_375 Ай бұрын

X = -1,0 rakhe to equation satisfy bhi to nahi ho raha hai

@Sudhanshuranjan2809 Ай бұрын

@@Aadi_375 bhai wo -1 ,0 ko solution nhi bol rha, wo bol raha(-1,0) k beech me ek solution hai aur (1,2) k beech

@jayantijoshi708 26 күн бұрын

I was drawing graph and thought that there are two solutions but suddenly remembered that an exponential function has positive base always, so those who say that there are two solutions are wrong...

@NileshArt916 Ай бұрын

Sir 11 ki kuch playlist nhi mil rahi hai kya karu? Help

@PiyushKanswal3769 Ай бұрын

Aap bahut acha samajhate ho

@samyakwagde4522 Ай бұрын

Aap accha samjhate hain❤❤❤🎉

@Ygsj653 Ай бұрын

I taught some new thing of index in this video. Thank you sir.

@SANJEEV-po3im Ай бұрын

Sir kya aap engineering math's ki new series start karwa sakte ho kya .apka samajh aata hai bus .😊

@vishnumehra0707 9 күн бұрын

Thank u sir ji 👍🏻👍🏻💯💯💯

@warsiboymotkee9789 28 күн бұрын

Sir ye log se nhi ho sakta kya

@Love_seeker-143 Ай бұрын

Sir iss question 3rd step tak toh maine bana liya tha but uske baad jo apne divide and multiply kiya hai woh mai nahi soch paya 😢. Apne ye thought kaise arise kiya aur 4 se kyu aur kaise apne predict Kiya pls tell sir.

@CatalpaGG Ай бұрын

Approach questions seh hi build hoti h

@Love_seeker-143 Ай бұрын

@@CatalpaGG woh mujhe pata hai but ye approach jo ki comparison ka hai maine bhi lagya tha but 4 sir ne jo 4 step mein divide multiply kiya hai woh literally critical thought tha.

@zavierbanerjee5171 Ай бұрын

You may ask this question : what number we can multiply with 64 and also raise it to power of 4 will give us the same number? Like 64*4 = 256 and (4)^4 = 256! This is because, in the equation, we want to bring it the x^(1/x) format... So, in the LHS, i.e, 4^(1/64), we multiply and divide by 4, so now the LHS => 4^(4/64*4) this is same as => (4^4)^(1/256) => 256^(1/256) !

@OmniCode786 Ай бұрын

4^x=x⁶⁴ Ln4^x =lnx⁶⁴ Xln4=64lnx Diff wrt x As ln4 and 64 are both constant Ln4d/dx x=64 d/dx lnx Ln4(1)=64/x X=64/ln4 X=46.16

@user-wu1yj9si8p Ай бұрын

Vha god of mathematics❤

@JoaoPedro-js6mk Ай бұрын

Guruji ek sawal hai... Kya aap complex or trigo ko ek sath krke kuch sawal Kara sakte ho... Kyu ki hame smjh me nahi ata kya kaha lagaye

@notstorm Ай бұрын

sir rough graph se to ham directly bol sakte hai 3 solutions honge, usme se ek x = 256 pe aayega?

@ShadowGaming-qf6tf Ай бұрын

BY COMPARING POWER X=64

@shlokthosar8094 Ай бұрын

Sir please start any series for Olympiads

@sentishiva29878 Ай бұрын

Here are the two solution which one is 256 And one between (1and 2) real solution

@user-tl1ot6st4j Ай бұрын

Sir function ki playlist gyab hoagye sir And ap ki 10 initial video bhi nahy hy sir And 1 to 11 nahy hy 12 to 17 hyy We are 2waiting please sir re uploaded

@Renu.Pahadi Ай бұрын

Sir your batch is awesome

@amits4744 Ай бұрын

I am already 30+ and not a student, but looks like this channel can become like 3Blue1Brown and BlackPenRedPen from India

@zavierbanerjee5171 Ай бұрын

Very nice approach! If you take the equality : 64*4 = 4^4 this is same as 32*8 = 2^8, then we get another question to ask : Solve : 2^x = x^32, which also can be solved on similar line...

@sandeepchakrwarti2468 20 күн бұрын

Thoda details mai kiya karo sir ji..ye divide mai kaise aaya

@09_i_tushargupta98 Ай бұрын

Calculus se bhi ho jayega Log le ke zyada behtar hai Itna thodi karna hota hai.

@SohamGhosh-xu9qk Ай бұрын

Did by graph 4^x is increasing function wheres x^64 is decreasing so it meets at one point 0,256

@Vr24j Ай бұрын

Did u used any website to draw these graphs 🧐🤔

@SohamGhosh-xu9qk Ай бұрын

@@Vr24j Use desmos

@kunjdubey2982 26 күн бұрын

Abe galat h 1.02 and -.9 hoga

@alankritgaming8254 Ай бұрын

Can we solve it by taking log both sides ❤

@rudrakshthaokar9789 Ай бұрын

Yes

@user-ox5gp3it6l Ай бұрын

You are genius sir❤

@Goodbro0007 Ай бұрын

Sir graph se answer alag aarahe hai

@Hkmotivat913 Ай бұрын

Sir class 11th batch KZfaq par digiye please nahi to aap par 500💵me class dijiye

@yuvrajkumar3961 Ай бұрын

Sir i was able to solve this question in a different way but quite similar I just multiplied the power by 1/64 then from RHS i analyzed that x=256 But sir i am unable to understand why some people are saying that it has 3 solutions pls explain sir

@Sam.69. Ай бұрын

X= 1/64

@motivationmastery5442 Ай бұрын

Yai question graph se ho sakta hai kya ?

@Idk2007-fx4cs Ай бұрын

Use graph to find number of solutions

@Oppositeofmyprofilephoto Ай бұрын

can we solve this with Logarithm?

@MrinalThakur-xi9tm Ай бұрын

Sir this was quite good problem

@ankeshiitkanpur2008 Ай бұрын

Can anyone please explain me the first step how sir have written 4= x^64/x

@Beliver_of_unseen Ай бұрын

Taking 64th root

@ankeshiitkanpur2008 Ай бұрын

@@Beliver_of_unseen ok thanks 👍

@tdstudymotivation Ай бұрын

Aap basic mathematics ki videos bna do sir

@ridul 24 күн бұрын

Every Olympiad students know these questions 😂😂😂😂

@mohdanas5688 Ай бұрын

Amazing sir ❤❤

@Love_seeker-143 Ай бұрын

4^x=x^64 o.t.l.b x.log4=64.logx X/64=log(x)/log4, X/64=log of x on base 4 (by base change prop.), X=64.log of x on base 4, O.D.B.S 1=64.1/(X.ln(4)), X=64/ln(4) sir kya ye tarika sahi hai pls tell. Pls pls

@pranavphiske3001 Ай бұрын

Bhai Maine bhi aise hi solve Kiya hai

@Love_seeker-143 Ай бұрын

@@pranavphiske3001 toh ye aise solve karna chahiye kya ya galat hai

@shivx3295 Ай бұрын

Galat hai aur dusri baat yeh solution nhi hai Bhai log dhang se padhle

@Love_seeker-143 Ай бұрын

Tujhe log galat dikha kya🤣 kyuki shayad tune log prop. Yaad na hai but Hain ye galat hai cause O.D.B.S is not valid for such equations and th value I have presented is wrong. But by graph maine dekha iska value jo hai vo .25 aur 1 ke beach mein intersect kar ra ha tha but accurate value nahi bata sakta🥲

@someone-jl4sj Ай бұрын

@@Love_seeker-143 What is O.D.B.S.

@Class-ld1on Ай бұрын

Sir ji aap ki sabhi video usefull hai❤❤ I like them Please make video on ramanujan conjecture 🎉

@Artist_Entertainment_Art 21 күн бұрын

Sir isko log ke rule ka use karke solve kr skte h

@user-fg1ii8wc3x Ай бұрын

why can't log be used sir?

@zeroplays9915 Ай бұрын

then what youll have to use Lambert W function which is not in jee syllabus

@housewiring1136 Ай бұрын

X=2 ka power 1/32 hoga

@Sugar-wx8qt Ай бұрын

I will suggest you to revise calculus😂😂 cause the graph will provide you 3 unique solutions for this equation..

@vandanarajpoot5544 Ай бұрын

गुरुदेव ITF की प्लेलिस्ट नहीं मिल रही है कहां गई आपने प्राइवेट🎉 तो नहीं कर दी है या फिर डीलीट तो नहीं हो गई यूट्यूब से। Please sir provide ITF playlist.

@NoName069 Ай бұрын

arey bhrata itna hindi😅

@vandanarajpoot5544 Ай бұрын

@@NoName069 I.e.

@vandanarajpoot5544 Ай бұрын

@@NoName069 what do you want say to me?

@mrinalkumardas7974 Ай бұрын

Just take logarithm both sides it will be solved easily.

@zeroplays9915 Ай бұрын

then what youll have to use Lambert W function which is not in jee syllabus

@pro_gamer_yt_6126 Ай бұрын

we can solve easily by taking log base 4 both sides

@samratahire1790 Ай бұрын

Thank you sir 😊❤😊

@user-sl9kd9wv2k 16 күн бұрын

yes

@mainakjana6903 Ай бұрын

By graphical method x=1.022 and x=-.979

@zeroplays9915 Ай бұрын

graph of a^x?

@mainakjana6903 Ай бұрын

@@zeroplays9915 same as e^x; little bit scaled

@railwaymathsspecial 15 күн бұрын

मैं तो orally solve कर दिया कुछ ढंग का question लाओ

@shubh2s305 Ай бұрын

Sir i did it by taking log on both side

@mr.sensei1502 Ай бұрын

Much useful

@NDA_Success_Tips Ай бұрын

Sir apke channel per video hide हो gai है 😢

@crazymathematician88 Ай бұрын

Take log both side of base 4 done in minf

@zeroplays9915 Ай бұрын

then what youll have to use Lambert W function which is not in jee syllabus

@crazymathematician88 Ай бұрын

@@zeroplays9915 what Is lambert w function? Can you explain me

@Rana_sir Ай бұрын

Phli baat to ye aapko achanak se nhi mila hai.

@Study77-lr3xm Ай бұрын

Maine ise log leke solve kar diya

@kashafarman3142 Ай бұрын

Sir what was that question

@ShershahSingh-hp1gc Ай бұрын

Is x = 3 is correct

@ajeetanand96 Ай бұрын

Sir there will be another two values according to desmos

@ketanverma3721 Ай бұрын

Sir i assumed x as 4 to the power m and then solve it❤

@Salim_Subzar 17 күн бұрын

POV : You are a Coder 💀 x ≈ 1.0223929402057803 x ≈ -0.9790169349577846

@omnatekar383 Ай бұрын

I tried on calculator but 256 is not answer I did 4(256)-256⁶⁴=0 But answer is not getting

@prateekanchal3115 Ай бұрын

Sir kya both side log with base 2 lekar solve nhi kiya ja sakata hai

@Naman_JEE25 Ай бұрын

Nahi bro solve nahi ho paa Raha

@amirkhursheeddar Ай бұрын

Why not to apply log on both side

@zeroplays9915 Ай бұрын

then what youll have to use Lambert W function which is not in jee syllabus

@rajjaiswal2379 28 күн бұрын

Question easy hai

@ab9555ab Ай бұрын

3 is the answer

@ayushverma6146 Ай бұрын

Done by using log base 4

@JEEASPIRANT2026-h2m 29 күн бұрын

Impossible 😂

@Prathamesh_Laddhad Ай бұрын

BY LOG

@princeraj7223 Ай бұрын

Sir log kyu nahi le sakte hai please 🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻 explain 😢😢😢😢

@Debarghya_ Ай бұрын

It has three real solutions as per Wolfram alpha

@user-bm9mn7ie7u Ай бұрын

whyy cant v take log

@rachitbajaj5301 Ай бұрын

Idhar log kyu nhi le sakte?

@spaceodyssey9788 Ай бұрын

X must be greater than 0. It's not given here. I think that may be the reason

@Gyan-fx9zx Ай бұрын

There are two more solutions 1.022, -0.979(approximately)

@zrpq Ай бұрын

How?

@Gyan-fx9zx Ай бұрын

@@zrpq well if you want to slove it alebraically you will have to use lambert W function otherwise just see the graph.

@zrpq Ай бұрын

@@Gyan-fx9zx 1 solution is clearly visible in graph in 1st quadrant as well as 2nd quadrant but 2nd solution in 1st quadrant is hard to find on graph.

@shivx3295 Ай бұрын

@@Gyan-fx9zx bisection method se bhi nikal jayega