Question

Hard One:

Answer 1

Find the real numbers x and y which satifies the equality:
\[|\ln(x^2+y^2+3)-2\ln(x+y-1)|+(9^x-4*3^{(x+y)}+3^{2y+1})^2=0\]

Answer 2

Probably no solutions.\[|\cdots | = - (\cdots)^2\]\[\implies |\cdots | = \rm a ~negative ~number\]

Answer 3

Unless you have the perfect square = 0, then it's a possibility.

Answer 4

PK only if
|...|=0 and (..)=0

Answer 5

*headach* T_T

Answer 6

If that perfect square term is zero, then \(-\rm perfect~square~term = 0\).

You now have to solve the equation \(\ln(x^2 + y^2 + 3) + 2\ln(x + y - 1) = 0\) and also check whether plugging in that value of \(x,y\) will give you the perfect square term \(=0\)

Answer 7

Another, and an easier way to see it is,\[|x | + y^2 = 0\]Only holds when both terms are zero since both terms are non-negative.

Answer 8

@BSwan Yup.

Not sure - is OS acting slow for you too?

Answer 9

What's the actual order of these replies?

Answer 10

ok, I'm leaving. OS is acting weird. I have to refresh a page thrice to submit one reply.

Answer 11

yeah its laging for me as well

Answer 12

so we have x+y>1

x^2+y^2=e^2(x+y-1)

3^2x +3^(2y+1) -4.3^(x+y)=0

Answer 13

@ParthKohli Are you sure that are no solutions here

Answer 14

did anyone try wolfram

Answer 15

Actually i found the solution without using wolfram

Answer 16

you can simplify the log expression

Answer 17

the key of the exercise it's in the log