24wow...nice discussion on a simple ques [1][1]
thats the USP of tiit [4][4]
18 Answers
24
62but karna.. the imporant thing here is that the powers are not integers..
If I asked you what are the integer solutions, then your answer would be correct..
that too only if you took +ve integers.. because for zero and non negative4 integers, the odd even thing wont work.. since 2^n wont be an integer any more (a small change in argument will actually lead to no solutions there as well)
but the point that I am trying to drive in is that you need both sides to be integers..
but here it is not the case.. You can have non integral solutions..
Liek for example...
no of solutions of 2x=1x
you could jump to the same conclusion because RHS is odd while LHS is even...
But you can see that x=0 is a solution..
21sir tats toh ok that we cant equate power of 2 which is in lhs with power of 3 which is in rhs
but why cant we do like wat asish has done above...i.e equating power of 2 which is in lhs with power of 2 which is in rhs and similarly for 3
1no sir ..one side is even and other is odd........so we cant equate......[92]
62see what you are saying will work if we were to find integral solutions or somethign like that..
otherwise,
answer a simpler question..
2x=3x+1
Will you get the answer by equating the powers?
62no it will not work out..
and the reason is that 2x.3y (The bivariate function) is not one on onto
106i\; meant \; 3^1.2^1 = 3^x.2^\frac{3x}{x+2}
so, \; x=1
equating x=1 and x=3x/(x+2)
but now i can see it leaves out some solutions
62you can find a couple of solutions by expressing -20 as the product of 4 integeers and hoping a solution with integer values of x
After that it is about factorization...
Otherwise you can observe the constant term here that will be 2484 and try some of the factors of this to get a root
Anotehr method would be to go for x2+15x+44 as t
so that the equation becomes
t(t+12)+20=0
solve for t.. then solve for x.
106i meant LHS coeff of 3 = RHS coeff of 3
and LHS coeff of 2 = RHS coeff of 2
106sir, cant we do directly by equating the coefficients of powers of 2 and 3 on both sides?
62Both these are from A Dasgupta... I have kind of been asked this around 10 times now :D
62this is same as
3^{x-1}=2^{1-3x/(x+2)}
Now solve it by taking log :)
