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[@1254v]

log(a^2)=2log(a) ❌ log(a^2)=2log(|a|)✅

[@AbcdAbcd-p5e]

Domain of f(x)=loga(x) ,a>0,a≠1 is (0,♾)

[@tuhinsubhraghosh9760]

Sir you are "Great" ....Because of you, I developed interest in mathematics...The way you approach the problems is really amazing...Keep teaching in the same way for years and years.........❤❤❤❤

[@aartibansal3899]

Sir i have paused the video at 10:00 and want to tell the mistake The mistake is that if power of y comes in front as the coefficient of logarithm then we need to put a modulus sign on y because as per question y can take negative values as well and argument of logarithm can't be negative. This is the problem in the solution.

[@ballPool-tj7nq]

Ye sirf even powers mai hoga na odd power mai to mod nhi aega ? Confirm kardo

[@vivekdubey8354]

@@ballPool-tj7nq nahi, odd powers me mod nahi ayega. ln (x^(2n)) = 2n.ln (|x|) ln (x^(2n+1)) = (2n+1).ln x

[@bshm_y]

Kash inka.... PYQ's course thoda sasta me mil jata

[@atharvsharma7648]

10:00 2 hatake mod y likhna prega, 8 values positive vali as well as negative vali side hogi and total solutions 16 honge...jus like x^2>p jab hota hai tab ya to a^2-b^2 se wavy curve lagado, ya direct mod x>sqrt(p) se evaluate krdo...

[@smitpatel5107]

Y ki power ko directly aage nai la sakte,kyuki y to negative bhi hai,given interval mein... Sir aap ne jab vo pehili baar step likha tha na,tab hi maine uss mistake ko identify kar liya tha... You are great sir ❤❤

[@arpitchauhan2147]

This concept is used used in one of the problems of black book... But i was not able to identify the mistake

[@NexInspireHub]

The Property You Applied, in which you put the power in front, i only valid when we can make sure that after shifting the power the log is still defined, but in this case when you put '2' in front at that time log y is not defined in [-10, 0). So we can't put the power in front, we have proceed it using mod or as it is

[@uguess3609]

Yesss same doubt.

[@NexInspireHub]

@@uguess3609 That's Not Doubt Bro, Sir asked where did i do wrong, i am telling him where he did wrong.

[@pleasejustlmb]

@@uguess3609 abey pagal h kya.

[@prajjawal4148]

Sir is solution me jb aapne y² me se 2 ko aage nikal liya tha y ki jo kuch values negative ho skti thi uski possibility khatam ho gyi since tb agr y me negative values nhi dal skte since vo out of domain ho jayega... Mujhe to yahi galti dikhi sir...ab mai aage ka video dekhne ja rha hu...

[@shampa364]

The thumbnail though 💀

[@KushalKalita-yf3yd]

Someone solve it- (i) log of (x+y) to the base 3 plus log of (x-y) to the base 3 equal to 1. (ii)2^(y/x+x/y) equal to 32

[@Iamarealdiamond3]

​@@KushalKalita-yf3ydAre you mental? This is not a board exam in which if we solve any problem in easier way then we will lose marks. It is MCQ exam, it is numerical value exam. We can use any method, and graph method is the easiest. Why you think so much? Remember one thing, you will remember it well. Remember many, you will be confused.

[@KushalKalita-yf3yd]

I apologise you for my mistake ,you solve it which you think better

[@rudrajadhav679]

Great Sir 🤩🤩 Your Concepts are deeply strong

[@kdhd100]

I am sure NTA answer sheet will have the solution as 8.....

[@thyagarajtrt1411]

Than even that's wrong

[@darshanlal263]

Then the Indian teachers would have to rebel for this

[@ThorfinnBus]

Real bro.. and even if we challenge they will pretend they are correct. This year I lost 15 marks without making a mistake in one exam.

[@kdhd100]

NTA would argue that since the coaching book answer is 8 the answer is 8..... End of conversation.....

[@bhavdeepsingh..]

On first screen. When we apply the most common log identity. It should be mod of y instead of just y. So at last step there will be 16 integers

[@ssgamer5693]

The inequality will never be satisfied by exactly 1 value of x,if it is satisfied there will be infinitely many values of x satisfying that inequality.... Question ko teda karke pucha hai

[@kashyaptandel5212]

“Atleast”

[@SamarthPatil-my5mh]

Sir ive paused the video at 9:51 The main reason why its wrong because whenever we have even powers of a variable in logarithm, while using the property we will have to apply modulus to satisfy the domain of the logarithm😊

[@harshitverma7814]

10:03 sir mujhe lg rha h ki question mein y^2 hai to y ki jo values positive mein hai vahi negative mein bhi valid hongi

[@Gopi12392]

Solve this pls in next video: 1)Prove that: (sinA.sin2A+sin3A.sin6A)/(sinA.cos2A+sin3A.cos6A)=(3tanA+2tan^2A+2tan^3A-6tan^4A-tan^5A)/(1-2tan^2A-6tan^3A-3tan^4A+6tan^5A)

[@himanshmehta3922]

Sl loney 😅

[@msharma6424]

Sir aapse questions kaise puch sakte h

[@SnehaRay-q8b]

VOTE FOR LECTURES ON NUMBER THEORY FOR ISI/CMI 👇👇👇

[@AmlanSarkar-wr2pr]

I also need the same bro as I am an ISI/CMI aspirant.

[@utkarshpal1792]

Yes I also need

[@Benzene42]

Ye hai kya

[@AmlanSarkar-wr2pr]

@@Benzene42 Bhai Kabhi bhi Indian Statistical Institute ka naam suna hai aur Chennai Mathematical Institute ka naam? Jaan lo a6 se taab pata chalega kya cheez hai I can guarantee these 2 institutes>>>>>>>>>>>>>>>>>IIT Bombay.

[@KPISEASON353]

​@@AmlanSarkar-wr2prmat batao bhai in chomuo ko, nhi toh befaltu me competition badh jayega

[@Jee2025-oz1zs]

Sir iota = iota power 4/4 Iota =1power 1/4 Iota = 1 How is it possible sir i just thought of this when I was doing revision of complex no Where is the catch??????????????

[@Root1729]

Mistake is that 1^1/4 is not 1 here

[@Jee2025-oz1zs]

@@Root1729 why??

[@Root1729]

@@Jee2025-oz1zs there will be 4 values and only iota satisfy , learn complex numbers bro then ask doubts

[@adstudypoint]

​​@@Jee2025-oz1zs bcoz. In x^4=1. Has four roots 1,-1,I,-i

[@tanmaykulkarni1540]

x^(1/4)=1 has 4 roots 1,-1,i,-i

[@shlokthosar8094]

Sir please start any series for Olympiads ..

[@dhairyagajjar8899]

Sir i did it by perfect square And then got the ans

[@Brown_Nikola.369]

Yes sir ISI number theory classes TOMATO book solutions please🙏

[@VishalKumar-ep5oe]

Bhaiya class 12th ka AOD CHAPTER KA. 6 unbelievable hidden lecture kaha par milega. Very need.

[@akhileshvijayvargiya4259]

Sir jab 2 ko aage lekar jayenga tab mode y aayega this is the mistake in 1st solution

[@rakeshranjan591]

sir i have a doubt please solve it . Sir log is only defined for positive numbers to why will we put mod

[@RajeevNayanPal]

Sir y2 me y ko positive hona zaruri nahi tha par jab sirf y hai to y ka positive hona zaruri hai ye condition Q me nahi thi hamne dali

[@RajKumar-uq1xp]

Sir log x power even number comes out to be log of mod x and because in this case x can be positive or negative

[@dalipprajapat1203]

sir log mein -ve define hi nahi hai to mod lagane knha jarurat hai

[@haggupaggu-dl6lx]

Thank you sir. Aaj ek nayi cheez seekhi

[@kavyagoyal6778]

We can't change the domain of function by applying modification to original situation

[@Education-35]

Essa too kabhi kisi ne padaya hee nahi sir Mene solve kiya pause karke too 8 hee aa raha tha But aage se galti nahi hogi

[@RAMKUMAR--------------696]

Sir very good explaination🎉🎉🎉

[@sidharthsingh9219]

y ke square ko remove karke hamne use positive assume kar liye aur negative values ko not aaceptable assume kar liya

[@shubham-z1l]

first inequality mai function is negative should'nt the D be negative then because the fucntion is never zero and always negative? but maybe im wrong kyuki it is not mentioned for all x edit : i just realised that vice versa of " if discriminant is negative then function is above or below the axis" is not true

[@supritykumari6989]

Unacademy ki beizzati

[@million-jv8dy]

But it is allen test question, the student who asked has commented 🎉🎉yet

[@HarshMehtani747]

Log y ki power property use karne ke baad wahan |y| aajayega

[@million-jv8dy]

THANKYOU SIR FOR SOLVING THE QUESTION ASKED BY ME , BTW IM IN ALLEN KOTA , ITS THE QUESTION OF ALLEN KOTA LOWER BATCH MINOR TEST Edit : YES , THE TEACHERS OF ALLEN KOTA PROVIDED US WRONG SOLUTION 😂🎉🎉🎉

[@SamridhSahshauray]

Sir apne y² ka domain toh liya hie nhi

[@AinurUlFijar.]

Log(a^n), n is even, then you have to put mod.

[@ch23satyendranathtripathi84]

sir jab log ko t put kiye the usme log decreasing function h

[@sumitdutta3311]

Sir aap iska solution bina **academy ka naam bataye bhi to de skte the aap ko to bas high light me aana hain saaf saaf dikh rha hain coatching ka naam unacademy hain 😅

[@lakshyadhimanld027]

ON TAKING POWER OUT, TAKEN CARE IS Y TO BE POSITIVE

[@Xyztaken50]

Y can be negative too

[@okiedokieyup926]

Sir log |y| (MOD aaega) aaega not log(y)

[@CHINMAYAKrishna-y5v]

Hurray muje samajh me aa gaya tha❤

[@rishabhjain2901]

Sir 1st sol mai aapne jab 2 ko bahar kra tab aapne anjane mai y ko positive assume kr liye

[@abhirupkundu2778]

Yea true, y must be positive

[@RAMBABU-wg2nj]

Before watching this lecture, my answer came 16.

[@mathswithsirsajid745]

Sir this is Sajid Ali from Pakistan Sindh province M.phil scholar in mathematics But I got so many concepts from you And with the help of your concepts Finally I passed my commission exam Recently working as subject specialist mathematics bps-17 in Pakistan 🙏🙏🙏

[@uttu2008]

Sir pls upload solution of question: sinπ/7. sin2π/7. sin3π/7.

[@agnibho2k8]

Already uploaded

[@shashwatorsamriddhithakur1774]

√7/8 is the answer right??

[@agnibho2k8]

@@uttu2008 (√7)/8

[@FlakernN]

​@@shashwatorsamriddhithakur1774you are wrong. Put π/7=0, so our answer comes as sin(0)sin(0)sin(0) = 0

[@06akshatnahar60]

​@@FlakernNBhai π/7 constant na ki variable uski value fixed hai. Thumara statement hi galat hai π/7=0=> π=0 Which is incorrect

[@Alexander0007]

Thanks a lot of sir ❤❤

[@rayquazabtc9131]

Sir maine bhi same glti kri😂😭

[@vishnulokiitmains137]

Sir ye common mistake bhi ho sakta hai Mai Manta hu aisi common nahi honi chahiye But kabhi kabhi ho jati hai So please don't create unnecessary problem Very good you are trying to make students correct But your thumbnail was very immature

[@AmanYadav-cg2kn]

sir y ke square ko aise hata nahi skate kyunki kya pata y negative ho, jo ki log mein possible nahi hai

[@NitheshGhorpade]

😂

[@kdhd100]

That is why we have to use the mod sign

[@AmanYadav-cg2kn]

@@kdhd100 are i know i just commented kyunki sir ne kaha tha

[@als2cents679]

Amazing !!

[@AdityaYadav-bf7it]

good solution....

[@dakshsehgal9847]

10:01 sir aapne square bahar lia so andar y pe mod lagana tha

[@imsparshyt]

exactlyyy

[@imsparshyt]

@@tiwary7320 are bhai because maan lo koi log 4 ko likh de log (-2)^2 to aise to log undefined ho jaega power directly bahar nikaal doge to... islie mod lgate hai taki always defined rhe log

[@imsparshyt]

@@tiwary7320 ehhh bro, log is not defined like this.... you can't have a non positive number inside log.... i suggest you first study the definition of log.... for short- log y base x means the power to which x should be raised to get y, so thus, and we always take x as positive .... (check the definition online)... thus, x to the power anthing can't come negative hence y cant be negative

[@imsparshyt]

@@tiwary7320 welcome

[@10.arnavchamorshikar55]

mistake identified and i solved this question correctly by myself that expression should be -2log(mod(y))

[@Jigneshiitian2025]

Sir due to you my maths is strong sir thanks sir

[@sidharthsingh9219]

Sir I solved this question in the first time

[@IIT572]

Sir I think this solution wrong bcz when log(m)^n applied property then the correct version is nlog(|m|).

[@Archit978]

Mode

[@keshavnagar1265]

Vha par y ki jagah mody aayega

[@vivekdubey8354]

ln (x^(2n)) = 2n.ln (|x|) ln (x^(2n+1)) = (2n+1).ln x

[@creatordip9900]

Mod of y

[@parellax]

i got 10 lmao

[@ARIJIT-g6j]

I cant assume 2logy = t..I think that's the mistake

[@akshat9188]

😂😂 bhai galti unhone bata di

[@kevinkamli609]

No you can

[@Arshadkhan73042]

Argument positive nahi kiya

[@Verma-b6d]

Sur apne 4 ko kat diya hai

[@creddym1]

Sir you are wrong

[@bhaktbabaaryan323]

No my brother sir is right Agar koi mistake ki hai to aap batao Mai apko suggest karta hu aap google par 'desmos' graph par iska graph bhi dekh sakte hai... Solution bilkul correct hai Ans:-16🙂🙂

[@AmbrishTiwari777]

r u from Unacademy?

[@bhaktbabaaryan323]

@@AmbrishTiwari777 no brother I do not prefer any online platforms