Sir yeh question maine just by equating the exponents mein kiya ho v gaya answer. I don't know if this process is correct or not but answer is coming -1,1,0 . I have tried the logarithmic process also but i did some mistake out there. Writing it at 0:30. Let's see the other methods.
Sir , i have an alternate and much easier approach, take x²= t and substitute in given equation to simplify powers. Then take log with base 2 on both sides of equation , and it will be converted to a simple cubic equation in t with root t=1 , which gives x= 1,-1 , not able to solve for x=0 but that can be checked by substitution in initial equation 🙂
@@RohitSingh-xq9ep bro RHS mein 2^x^4 +1 ko 2×2^x^4 likh do fir log lo , isse ho jayega , mera correct hai , hope ab aap samjh paye main kya kehna chahta hu
I solved this question by dividing 2 ki power x^2 and then LHS me 1 bna liya ,2 ki power something minus 2 ki power something ayega equal to 1 or ye tabhi satye h jab phele wala something equal to 1 ho or baad wala something equal to 0 ho vaha se x ki values ajayegi
Can we also solve the problem in a bit different way, we can take 2x^2=a , then equation will be a^3+a=2a^2,now bringing RHS to LHS , the equation will be (a2-a) *(a-1)=0 Now we have two solutions, a2=a,now substituting with 2x^2 , we get x^4=x^2 or x^2=1 thus x= +-1 Another solution a=1 gives solution x=0
Sir I have used this approach 1)Divide both LHS and RHS by 2^x² 2) Transpose all the terms except "1" to RHS 3) It turns out the only combination for the difference of two powers of "2" to be equal to 1 is 2^1 - 2^0 4) Compare on both sides and get the answers
simple trick of these type of questions - in AM and GM equality holds in all those questions in which we have find the proper value of the terms so equate both terms you will get the value of x This trick works on all these type of questions .🙂
@@asthronoxde8030 take 2 as any variable let's say a and x² as any variable let's say b. Take x⁴ as c this equation becomes a^bc + a^b = a^(c+1) -> a^bc + a^b = a^c × a Take a^c to other side (as division) a^b(c-1) + a^(b-c) = a Now substitute all the values 2^x²(x⁴-1) + 2^(x²-x⁴) Simplify and solve for zeroes and you will get all the answers
I figured the answer within a minute , cuz I remembered my past encounter with an exponential eq. q... I think it was from probability or pnc exercise (numerical type) of vikas gupta . My learning from it was that when two exponents of base 2 can be conveniently added or subtracted to give another exponent in same base ,i.e. 2 when their powers are same (in case of addition) and at a difference of 1 (when being subtracted , cuz then u get a 1/2 factor which can incorporated as -1 power) .... so i straightaway equated x2 and x6 , which gave meaningful answers , or else i would have further tried taking one of the terms to RHS and tried creating that difference of 1 Anyway , sir ur solution was also great , and I will try to approach with am gm also now in such q.s
Bro I solved it in a different way... I took 2 x2 as y and solved it as quadratic eq... But I only got 1 value which was 0.... so I was not sure about my solution
Ya question sach me badiya he Par sir at 6:03 jo aapne ( x⁶+x²)/2 ko x⁴ likh Diya . Vaha nahi samjha Bcoz vaha ( x⁶+x²)/2 >= x⁴ tha na ,so apne direct use use kese kiya ?
@@adarshranjan2707 arey mera matlab integral definite hai ya indefinite aur iota woh root -1 hai ya iota summation ka hai? sorry itne sawalon ke liye but aapka sawal clearly nahi diya hai...;( waise tough dikh raha hai
Sir I have seen this question and this is an amazing question of solving exponents , i have find a method to solve this question , so first take 2 ^x^4+1 to the denominator of 2^x^6 + 2^x^2 so we get 2^x^6+2^x^2/2^x^4+1 as 1 and we can now seprate it and as know the property of exponents we can write 2^x^6-x^4-1 + x^2-x^4-1 =1 , we know only x^6-x^4-1 and x^2-x^4-1 can have a single value of -1/2 to satisfy the equation , so then we can solve x^6-x^4-1 by simplifying it as (x^2)^3-(x^2)^2-1=-1 , let x^2 = t we can get two values of t as o and 1 and then substituting t as x^2 we can get x as o and +1or -1. Again thanks sir for coming with such type of nice questions .
Sir, if we take 2^x^2 be A Equation esi hogi A^3+A=A2×2 A3+A/A2=2 A2+1/A=2 Then quadratic equation ban jaygi A^2-2A+1=0 (A-1)(A-1) factor ho jayge toh A equal to 1 Humne 2^x^2 ko A let Kiya tha toh X^2=1...X=+ - 1 Y bhi toh shi hua😷
Very simple approch : if the base of exponents is same so we can let x^6+x^2=x^4+1 we can simplify it and = x^8-x^6+1=0 we can made it look like cubic by letting x^2 = n so n^3 - n^2+1=0 solve the cubic and the root will be the answer
It's also possible to solve it by let t=2^x^2. (t> or =1). The equation become: t^3+t=2t^2. Then we will have (t^2-t).(t-1)= 0. Hence t=1 or t^2=t. In case t = 1 we have x = 0. In case t^2 = t we have x^4 = x^2, hence x= 1 or -1.
Sir, I am glad that you have being such a amazing equation in the eq the first step was the toughest.plz in future it will be helpful that you can make a video on how to identify the application of am and gm in a perticular equation.
Just take 2 power x4 divide on both side and you will see that sum two positive no. Is coming so u can individually solve each of the terms by equating both of them to 1 and ans will come 0 1 -1
Sir I found using hit and trial that 0,1,-1 will be roots but the way you proved that no other roots will be possible using concept of am gm was op to power op
Sir mene aise Kiya iss se zero nahi aa rha baki do roots aa gaye kya ye taika sahi hai Dono taraf log base 2 se multiply kar diya uske baad log ki property use karke ye equation bani x^6-x^4+x^2-1=0 se roots aaye +,-1 Sir isme kya galti ki please batana🙏🙏🙏🙏🙏🙏🙏 reply me
Sir I used Fermat's Last theorem, to deduce that exponent will be either 2 or less than 2, and then applied some logic on it to find out possible values, by using fermat's last theorem, which states that if a equation is given as (X^n+Y^n=z^n) then the maximum possible value of n is 2, I calculated the first value as 1, and then i saw that powers are in even so i instantly understood that it will accept -1 as well, and i then saw that in rhs if we upt x=0 then we get 1+1 and in lhs if we put x=0 we get 2^1 which is 2, hence proving lhs=rhs
I have solved this question by using simple algebra and exponent's law.. . But at the last me and sir both got the same eqn , and the answer was also the same .. So, whether my method is correct or not ?
Sir can we do this question by comparing as in question 2power x power 6 is added to 2 power ze and we can also write the rhs as the sum of 2 power x power 4 multiple by 2 power 1 Sir pls tell whether my method is write or wrong although I got correct ans
I solved it different way. I am in 9th and didn't understood what was going on the video but see my solution 2^x⁶+2^x²=2x⁴+¹ let x²=t 2t³+2t=2t²+¹ 2t(2t²+1)=2t²×2--(1) On comparing the terms 2t=2t² t=t² t=±1--(2) From (1) 2t²+1=2 2t²=1 Which is possible when t²=0 t=0--(3) From (2) and (3) t=+1 t=-1 t=0
I just devided by 2 then I got 2^(x⁶-1)+2^(x²-1)=2^x⁴ from there I took 2^(x²-1) common from left side you might be wondering how (x⁶-1) Can be written as ( x²-1)(x⁴+x²+1) , after that I got 1^(x²+x⁴+1)+1=2^(x⁴-x²+1) then any power to 1 is one so you can happily write 1+1=2^(x⁴-x²+1) Then 1=2^(x⁴-x²) take log bot sides then you get (x⁴-x²)log2=0 from there you get x=0,±1
hello sir, mujhe aapse puchna tha ki maine jo approach try ki iss question mai, woh sahi hai ya galat. pehle LHS mai hum 2^x^2 common le sakta hai. therefore; 2^x^2(2^(x^6-x^2) + 1)=2^(x^4+1) now if we take 2^x^2 to RHS we get; 2^(x^6-x^2) +1=2^(x^4-x^2+1) which implies that 2^(x^4-x^2+1)-2^(x^6-x^2)=1; we also know that this is true only for 2^1-2^0. therefore; (x^4-x^2+1)=1 and (x^6-x^2)=0 which means that x= -1,0,1.
i had a question... pls answer this...... can we just first take 2^x^^2 common and then we will get ( 2^x^^2(2^x^^4+1)=2^x^^4*2^1) and we can get our answers by this step by comparing as i am in 9th and i have not done AM and GM
I did it in a more simpler way !! Russia Just take RHS in LHS by dividing.. Hence in LHS sum of two exponential terms and RHS=1.. This equation is possible only when the two powers in LHS are -1.. This gives us x=-1,0,1🎉🎉 Feeling good by finding a simpler solution 🎉🎉🎉
This also can be solved by making the a identity (a+b)² It is difficult to write here.....Sorry..... If photo option would be there then I can show u all...
5:40 complications start from here.
Itna karne ka jaroorat hi kya tha sir itna AM GM uske andar ghusne ki ...isko directly log Wale se v kar sakte the
Mai aapke jaisa Genius Ban ne ki kosish ker raha hu
Mai bhi ab doubt questions solved kerne laga apne youtube channel per sir
Sir yeh question maine just by equating the exponents mein kiya ho v gaya answer.
I don't know if this process is correct or not but answer is coming -1,1,0 . I have tried the logarithmic process also but i did some mistake out there.
Writing it at 0:30. Let's see the other methods.
Sir you have the spirit that every student wants in their teacher. We want more topic wise JEE mains and advance question discussion and explanation.
Hero Of Mathematical world 🌎.!
He Has Its Own Style Of Teaching Mathematics ..!!
We Can Say a Unique Style Of Teaching Mathematics.!!
Sir , i have an alternate and much easier approach, take x²= t and substitute in given equation to simplify powers. Then take log with base 2 on both sides of equation , and it will be converted to a simple cubic equation in t with root t=1 , which gives x= 1,-1 , not able to solve for x=0 but that can be checked by substitution in initial equation 🙂
Yes
Substitute 2^x^2 then this become more easier
Bro your method is wrong log ( 2^t^3 + 2^t) = (t^2 + 1)log 2
Ab batao isse kaise solve karogai
log ( a+b)= Log a + log b hota nhi hai
@@Mohit_Choudhary._ can't do that because 2^x^6 hai jo 2^x^2 ka cube nahi hota , meine bhi pehle wohi socha tha
@@RohitSingh-xq9ep bro RHS mein 2^x^4 +1 ko 2×2^x^4 likh do fir log lo , isse ho jayega , mera correct hai , hope ab aap samjh paye main kya kehna chahta hu
in this type of question always first try hit and trial method!! it worked here! 1 and -1 and 0 are most common numbers for hit and trial
I solved this question by dividing 2 ki power x^2 and then LHS me 1 bna liya ,2 ki power something minus 2 ki power something ayega equal to 1 or ye tabhi satye h jab phele wala something equal to 1 ho or baad wala something equal to 0 ho vaha se x ki values ajayegi
Sir please bring questions from AIME , Canadian MO , Bulgarian MO and IOQM
Titu books too 😁
Can we also solve the problem in a bit different way, we can take 2x^2=a , then equation will be a^3+a=2a^2,now bringing RHS to LHS , the equation will be (a2-a) *(a-1)=0
Now we have two solutions,
a2=a,now substituting with 2x^2 , we get x^4=x^2 or x^2=1 thus x= +-1
Another solution a=1 gives solution x=0
Sir I have used this approach
1)Divide both LHS and RHS by 2^x²
2) Transpose all the terms except "1" to RHS
3) It turns out the only combination for the difference of two powers of "2" to be equal to 1 is 2^1 - 2^0
4) Compare on both sides and get the answers
simple trick of these type of questions - in AM and GM equality holds in all those questions in which we have find the proper value of the terms so equate both terms you will get the value of x
This trick works on all these type of questions .🙂
Solution karke btana bro
Did it using the basic exponential laws without using AM GM inequality. A smart 10th grade student can surely do this too. Wonderful question.
Can u tell me how?
@@asthronoxde8030 take 2 as any variable let's say a and x² as any variable let's say b. Take x⁴ as c
this equation becomes
a^bc + a^b = a^(c+1)
-> a^bc + a^b = a^c × a
Take a^c to other side (as division)
a^b(c-1) + a^(b-c) = a
Now substitute all the values
2^x²(x⁴-1) + 2^(x²-x⁴)
Simplify and solve for zeroes and you will get all the answers
It is wrong
th-cam.com/video/r21HrxrtBrE/w-d-xo.html
Why 10th grade 8th mein hoon I tried two ways one by the help of log and another by substituting 2^x^2 as y
I figured the answer within a minute , cuz I remembered my past encounter with an exponential eq. q... I think it was from probability or pnc exercise (numerical type) of vikas gupta . My learning from it was that when two exponents of base 2 can be conveniently added or subtracted to give another exponent in same base ,i.e. 2 when their powers are same (in case of addition) and at a difference of 1 (when being subtracted , cuz then u get a 1/2 factor which can incorporated as -1 power) .... so i straightaway equated x2 and x6 , which gave meaningful answers , or else i would have further tried taking one of the terms to RHS and tried creating that difference of 1
Anyway , sir ur solution was also great , and I will try to approach with am gm also now in such q.s
Bro I solved it in a different way... I took 2 x2 as y and solved it as quadratic eq... But I only got 1 value which was 0.... so I was not sure about my solution
@@taksh8989 you won't be able to substitute that way as 2^x^6 isn't y³
I tried and did it in first attempt... felt great, thank you Aman Sir for such videos!!❤
ye just put 2^(x^(2)) = t and solve cubic very easy
th-cam.com/video/r21HrxrtBrE/w-d-xo.html
@@GaurangAgrawal0exactly
What about +- i (iota) question mein kaha bola hai ki real solution
First step taking both side log then log 2 cancles easch other then common x² and cancel the bracket x²=1 , x=±1 simple
Ya question sach me badiya he
Par sir at 6:03 jo aapne ( x⁶+x²)/2 ko x⁴ likh Diya . Vaha nahi samjha
Bcoz vaha ( x⁶+x²)/2 >= x⁴ tha na ,so apne direct use use kese kiya ?
Sir i have a question for you
9999
Lim integeration ( summation ([x]^2)/x^(e+i))dx
X-->infinity
Please solve this problem.
bhai integration ke limits kya hai? aur kya woh greatest integer functinon hai kya? aur is e+i ya e+1?
Bhai wo iota hai aur intgrn ka limit hai ,wo GIF ka sign hai[]
@@adarshranjan2707 arey mera matlab integral definite hai ya indefinite aur iota woh root -1 hai ya iota summation ka hai? sorry itne sawalon ke liye but aapka sawal clearly nahi diya hai...;( waise tough dikh raha hai
Indefnite hai bhai
Summation X ke liye hai
Sir please solve pells equation
Y²-nX²=1 find n&x&u
Sir I have seen this question and this is an amazing question of solving exponents , i have find a method to solve this question , so first take 2 ^x^4+1 to the denominator of 2^x^6 + 2^x^2 so we get 2^x^6+2^x^2/2^x^4+1 as 1 and we can now seprate it and as know the property of exponents we can write 2^x^6-x^4-1 + x^2-x^4-1 =1 , we know only x^6-x^4-1 and x^2-x^4-1 can have a single value of -1/2 to satisfy the equation , so then we can solve x^6-x^4-1 by simplifying it as (x^2)^3-(x^2)^2-1=-1 , let x^2 = t we can get two values of t as o and 1 and then substituting t as x^2 we can get x as o and +1or -1. Again thanks sir for coming with such type of nice questions .
Could've also been done by taking log 2 on both sides and assuming x² to be y and then solving the cubic eqn.
That's how I did it
Howw log(a+b) is not log a+log b
Sir, if we take 2^x^2 be A
Equation esi hogi
A^3+A=A2×2
A3+A/A2=2
A2+1/A=2
Then quadratic equation ban jaygi
A^2-2A+1=0
(A-1)(A-1) factor ho jayge toh A equal to 1
Humne 2^x^2 ko A let Kiya tha toh
X^2=1...X=+ - 1
Y bhi toh shi hua😷
Done it by simple exponent steps from the concept of class 7
Very simple approch : if the base of exponents is same so we can let x^6+x^2=x^4+1 we can simplify it and = x^8-x^6+1=0 we can made it look like cubic by letting x^2 = n so n^3 - n^2+1=0 solve the cubic and the root will be the answer
Sir plise make videos for 10th mooving 11th students I am your great fan I love yo sir
Bhannat sir never disappoint us...😃😃😃
th-cam.com/video/r21HrxrtBrE/w-d-xo.html
Sir but how do we get to know that we have to apply am gm 🤔🤔please tell
Sir ye black book ka question hai , 2 ke jagah pe 16 hai
Sir aapse humari bhot help hoti hai
Sir please continue such videos
As I love maths
It's also possible to solve it by let t=2^x^2. (t> or =1). The equation become: t^3+t=2t^2. Then we will have (t^2-t).(t-1)= 0. Hence t=1 or t^2=t. In case t = 1 we have x = 0. In case t^2 = t we have x^4 = x^2, hence x= 1 or -1.
That's wrong
Sir, I am glad that you have being such a amazing equation in the eq the first step was the toughest.plz in future it will be helpful that you can make a video on how to identify the application of am and gm in a perticular equation.
It can be solved without am GM also
Just take 2 power x4 divide on both side and you will see that sum two positive no. Is coming so u can individually solve each of the terms by equating both of them to 1 and ans will come 0 1 -1
Put x^2=t and adjust the equation and I found it easily t=0 and 1 is the solution so x=1,-1,0
kaise bhai
0:43 aapke ye kehne pe mere dimaag mei achanak se click hua ki x =0 toh nahi hai kahi. aur wwahi hua
Only channel which never wastes spent time
Sir you are great teacher of mathematics .
Sir I have a problem of limit
Limit x tends to zero (X^X)
It ha direct formula please watch sir limit video lectur
😍😍sir.....kya sir mile hai mujhe
Bas itna kehena chahunga 1 dum bhannat😎
Very interesting sir.
Please provide daily this type of video
Sir I found using hit and trial that 0,1,-1 will be roots but the way you proved that no other roots will be possible using concept of am gm was op to power op
Best method, as if you see the eqn, the soln can be arrived by trial method putting x=0,1and -1
Kya mujhe koi example deke samjha sakta hai ki ye am gm inequality kaha aur kab use karte hain
This is kind of a well-built puzzle using inequality, I guess. But I found the solution brilliant.
Sir root of -1(i) will be another value of x for the equation
Nahi hoga LHS=1 and RHS=4..
I solved it without any problem thanks sir
Method yahi hai .ya different hai
@@Aghtejewgbs Different hain
Sir please bring more problems on trigonometry as it is feared by most of the students
Sir u made me fall in luv with mathematics ✨🌼
Overall mujhe aapka solution kafi pasand hai
sir how do know that here we were suppose to used AM - GM, I am just confuse about where we need to use it exactly
Sir mene aise Kiya iss se zero nahi aa rha baki do roots aa gaye kya ye taika sahi hai
Dono taraf log base 2 se multiply kar diya uske baad log ki property use karke ye equation bani x^6-x^4+x^2-1=0 se roots aaye +,-1
Sir isme kya galti ki please batana🙏🙏🙏🙏🙏🙏🙏 reply me
Sir I used Fermat's Last theorem, to deduce that exponent will be either 2 or less than 2, and then applied some logic on it to find out possible values, by using fermat's last theorem, which states that if a equation is given as (X^n+Y^n=z^n) then the maximum possible value of n is 2, I calculated the first value as 1, and then i saw that powers are in even so i instantly understood that it will accept -1 as well, and i then saw that in rhs if we upt x=0 then we get 1+1 and in lhs if we put x=0 we get 2^1 which is 2, hence proving lhs=rhs
isnt that for integers only?
Sir i took x^2 as y and substituted and solved. Got the same answer with 0 help from your solution for the 1st time.
Same 😁 x^2=t
I have solved this question by using simple algebra and exponent's law.. .
But at the last me and sir both got the same eqn , and the answer was also the same ..
So, whether my method is correct or not ?
Yup
Concept is clear from yr. Video, thanks sir.
Sir would plus/minus i will also be two additional solutions?
Every time i one your video sir i get new concept and knowledge
Sir can we do this question by comparing as in question 2power x power 6 is added to 2 power ze and we can also write the rhs as the sum of 2 power x power 4 multiple by 2 power 1
Sir pls tell whether my method is write or wrong although I got correct ans
Yes, bro. Ofc
I solved it different way. I am in 9th and didn't understood what was going on the video but see my solution
2^x⁶+2^x²=2x⁴+¹ let x²=t
2t³+2t=2t²+¹
2t(2t²+1)=2t²×2--(1)
On comparing the terms
2t=2t²
t=t²
t=±1--(2)
From (1)
2t²+1=2
2t²=1
Which is possible when t²=0
t=0--(3)
From (2) and (3)
t=+1
t=-1
t=0
Arr bhai nhi Galat hai.
Second step ma 2^t^3 hoga similary 2^t=2^t^2+1 hoga.
So third step ma common nhi aayega
Are bhai itna dimag mat lagao tum abhi 😂 😂
Wrong solution 👍
power kaise common liye bro 't' was in power
ye sab chodo bhai tum log prmo/rmo ki taiyaari kro , sahi mauka hai (for the 9th class guy here) , I am in 12th now , mujhe to pta hi nahi tha 9th me
Please start jee advanced previous years question series
Sir, you are brilliant and your teaching ability is too good and very easy to understand.
I just devided by 2 then I got 2^(x⁶-1)+2^(x²-1)=2^x⁴ from there I took 2^(x²-1) common from left side you might be wondering how (x⁶-1)
Can be written as ( x²-1)(x⁴+x²+1) , after that I got 1^(x²+x⁴+1)+1=2^(x⁴-x²+1) then any power to 1 is one so you can happily write 1+1=2^(x⁴-x²+1)
Then 1=2^(x⁴-x²) take log bot sides then you get (x⁴-x²)log2=0 from there you get x=0,±1
Sir I Have A Question For You Please Amake A Video On This Question’s Solution :- Find All Integer a^p=b!+p
Sir iss problem me dono side log ka multiply krke bhi solve Kiya ja skta he na..
Please share the details of your board for writing.
Sir Taylor aur maclaurin series pe ek detailed video banaiye please... Unke proof,examples ke saath.
How u considered am and gm
Kiya kru achi kuch samajh nhi aa rha aap scilab practical kar baye na bscit ka please sir
Cant we take log of all three and then devide all by log(2)
Both side log leke v solve kr sakte hai sir..
Add log on both sides the values of x are 1 and -1
Oh amazing concept thanks
hello sir,
mujhe aapse puchna tha ki maine jo approach try ki iss question mai, woh sahi hai ya galat.
pehle LHS mai hum 2^x^2 common le sakta hai.
therefore;
2^x^2(2^(x^6-x^2) + 1)=2^(x^4+1)
now if we take 2^x^2 to RHS we get;
2^(x^6-x^2) +1=2^(x^4-x^2+1)
which implies that
2^(x^4-x^2+1)-2^(x^6-x^2)=1;
we also know that this is true only for 2^1-2^0.
therefore;
(x^4-x^2+1)=1 and (x^6-x^2)=0
which means that x= -1,0,1.
Done in 2 min. but with different method but this method is awesome.
sir but what about +i and -i complex sol of above equation
Complex kidhr lagega?
Sir hum class 9th mein hain. Ye question humko samajh mein nhi aaya! kya yeh mere syllabus mein hain 🙄
AP starts in 10 not in 9
Sir yaha to 3 nmbr hai am gm nikalne ke Liye lekin aapne sirf 2 hi nmbr ka am gm inequaluty use Kiya hai
Last me x²(x³-1) ayega do answer 0,1
But answer mai -1 bhi hai. Somewhere mistake hai
Sir, answer of x^4 will be 4 answer, but you write only two why sir?
Thank you so much.
i had a question... pls answer this...... can we just first take 2^x^^2 common and then we will get ( 2^x^^2(2^x^^4+1)=2^x^^4*2^1) and we can get our answers by this step by comparing as i am in 9th and i have not done AM and GM
If u are in 9th ( in 10th I'm supposing right now) then there is no other way to solve this
Sir , west bengal class 12 th ka math suggestion need
🤣🤣🤣🤣
@@Pratimkoley8238 what the....... Pratim
Sir can we solve it by using sum of gp???
Sir could u solve more questions with exponents over exponents question? They confuse me a lot.
sir ye question badhiya tha
They are in AP
Apply log both sides and then you can solve the problem easily
Yes
Yes
Yes
I did it in a more simpler way !! Russia
Just take RHS in LHS by dividing..
Hence in LHS sum of two exponential terms and RHS=1..
This equation is possible only when the two powers in LHS are -1..
This gives us x=-1,0,1🎉🎉
Feeling good by finding a simpler solution 🎉🎉🎉
Sir hum isko AM GM se hee solve kiyu kar rhe hai???
Thank you sir
x = 0, substitution
Sir iss case me maine hit and trial method se kiya and easily ans aa gaya
Thanks sir ❤️❤️
bhut saal baad aapki vedio dekhii lekin question karne mein mazaa aaya # bhannatmaths
Sir Roz ese beautiful video banate jao❤️
I took log on both sides and solved for x, I got x = 1 and x = (-1)^1/3
Sir dimag ghumane wala problem tha lekin maja to bhaut aaya😊
Sir last m x power 6 ko x power 2 kyu kiya ??
Log lekar bhi solve kar sakte hai phir quadratic eqution ban jayega.
This also can be solved by making the a identity (a+b)²
It is difficult to write here.....Sorry.....
If photo option would be there then I can show u all...
Sir this question can also be done by taking 'log base 2' both sides
Noo