if we assume everything to be greater than 1 and x less than x1 x2 x3....xn then half term will open with positve and half will open with negative in LHS and all will open with positive in RHS (mod) after than just manupilate and we get x1.x2.x3.x4.x5......xn in a quadratic form with constant 0 since we assumed all to be greater than 1 then their multiplication cant be 0 but will tend to 1 or be 1 if we assume them to be >= 1 lekin ek problem hai ki baaki cases ka kya hoga toh my solution is inconsistent
If log(xx1) is negative then log(x/x1) is always positive and if log(x/x1) is positive then log(xx1) is always negative. whatever possibility is applied, both of these functions will break out of the mod in opposite signs unless they are both equal to 0. hence it's positive to say that |log(xx1)| +|log(x/x1)| = |log(xx1) - log(x/x1)| = |2log(x1)|. Once you solve this further, you can see that the only possible number is 1.
bhaiya mene is question ko iss tare ke se kia lag bhag same he par thoda sa long hai Step 1 : separate all the terms |logx+logx1|+|logx+logx2|+.............+|logx+logxn| + |logx-logx1|+|logx-logx2|+.............+|logx-logxn| = |logx1+logx2+....+logxn| Step 2 : Separate the equation in 2 cases and solve them Case1 logx+logx1+logx+logx2+.............+logx+logxn + logx-logx1+logx-logx2+.............+logx-logxn = nlogx+nlogx = logx^2n log(x)^2n=log(x1.x2.x3...xn) (x)^2n=(x1.x2.x3...xn) So this is only possible when all the numbers same equal to 1 Case2 logx+logx1+logx+logx2+.............+logx+logxn + logx1-logx+logx2-logx+.............+logxn-logx= log(x1.x2.x3...xn)+ log(x1.x2.x3...xn) = log(x1.x2.x3...xn)^2 log(x1.x2.x3...xn)^2 = log(x1.x2.x3...xn) (x1.x2.x3...xn)^2 = (x1.x2.x3...xn) And this is only possible when all the numbers same equal to 1 So answer is 1
In case 2 , you removed the modulus without - ve sign. And when (x)^2n=(x1.x2.x3...xn) [ not given that x=x1=x2=... =xn ] there will be infinite solutions. eg:- Let n=2, so there exist x, x1 and x2 only. Now x^4 = (x1.x2) is satisfied by x=2, x1=8 and x2= 2
did it vg sir also covered it in his classes not the exact question but the concept good question i after solving quite some questions on log i can say that this is not the most difficult one but yeah the most conceptual one but anyway it was fun to see you doing it love your channel btw keep it up ....
@@codexyz2887 check his apni kaksha lectures in one of them you will find him explaining these concepts after watching all letures you will understand that each and every concept was already covered in vg sirs class not direectly but yeah its same
Maine logxxi ko a mana Tha aur logx/xi ko b Tho idhar lal + lbl =l (a-b) l de rakha Tha question me, ab dono side square karke solve kare tho humko finally (lal+lbl)²=0 and this is possible only when lal=0 and lbl=0, x/xi=1 and xxi=1, this is only possible when x and xi both are 1 Iss method me property ka use bhi nahi hua Anything wrong in this method?
Mera bhi 1 aa gaya tha , lekin apne us solution ko galat bola tha . Kyunki mod me kaat nhi sakte , vhi solution 10 bacho ne or bheja tha😂 ( allen vale group me ) .
Ans: x=1 Approach: Using Logarithmic Properties We can Combine all logs on LHS as log{(x^2n)(x1x2x3....xn)/(x1x2x3...xn)} = log(x^2n) RHS can also be written as log(x1x2x3x4x5...xn) Now if we equate LHS = RHS After removing logs: x^2n=x1x2...xn From here we can infer x = 1 or 0 0 is neglected as it's not in the domain of Log(x) This is my approach.
thought of this approach initially solve karne baitha same logic same formula lagaya but further aage jaake confuse ho gaya ... my bad...will try better next time ...but got the solution instantly
bhaiya mene is question ko iss tare ke se kia lag bhag same he par thoda sa long hai Pin my answerrr plz Step 1 : separate all the terms |logx+logx1|+|logx+logx2|+.............+|logx+logxn| + |logx-logx1|+|logx-logx2|+.............+|logx-logxn| = |logx1+logx2+....+logxn| Step 2 : Separate the equation in 2 cases and solve them Case1 logx+logx1+logx+logx2+.............+logx+logxn + logx-logx1+logx-logx2+.............+logx-logxn = nlogx+nlogx = logx^2n log(x)^2n=log(x1.x2.x3...xn) (x)^2n=(x1.x2.x3...xn) So this is only possible when all the numbers same equal to 1 Case2 logx+logx1+logx+logx2+.............+logx+logxn + logx1-logx+logx2-logx+.............+logxn-logx= log(x1.x2.x3...xn)+ log(x1.x2.x3...xn) = log(x1.x2.x3...xn)^2 log(x1.x2.x3...xn)^2 = log(x1.x2.x3...xn) (x1.x2.x3...xn)^2 = (x1.x2.x3...xn) And this is only possible when all the numbers same equal to 1 So answer is 1
<a href="#" class="seekto" data-time="192">3:12</a> Bhaiya ek doubt tha ki jab apne yeah property lagai to sare terms mod ke ander plus me kese gye mtlb ki | log(x1) | + |logx2| hoga na
I like your videos a lot but in this particular video, there are a lot of false claims. You proved that sum of ln(xi) is 0 does not imply that all xi have to be 1.
So what should others do praise you on how great of a human you are or what.... Ok you solved it, so what its not some ground breaking achievement As the question was not that hard....
@@harshitkunjalwar i mean im sharing my opinion that its not tough . plus im preparing for smth Which is tougher than jee advance in grade 9(not flexing but want to say this since you commented Like this )
