if solved graphycaly what will the figure be?

Question

log with base (x^2+y^2) of (x+y) to be >1(greater than 1)

in my solution it's a square without the sides and with corners in(0;1),(1;0),(-1;0),(0;-1).But sb told me it was a cycle with radios sqrt1/8.Control my solution or give the right one...!!!Thank's!!!!!!!!!!!!!!!

Answer 1

log[base x^2+y^2] (x + y) > 1

If we take the antilog of both sides (where the base is (x^2 + y^2)), then we get

x + y > (x^2 + y^2)^1, or, quite simply,

x + y > x^2 + y^2

To determine what this figure would be, we solve the corresponding equality and then perform tests on what area to shade. That is, we want the graph of

x + y = x^2 + y^2

Moving everything to the right hand side,

0 = x^2 - x + y^2 - y

Now, completing the square for the x and y variables, this means adding a (1/4) for the x terms and (1/4) for the y terms. So we're adding (1/4) + (1/4) to both sides.

(1/4) + (1/4) = x^2 - x + (1/4) + y^2 - y + (1/4)

Factoring both of the square trinomials,

(1/4) + (1/4) = [x - (1/2)]^2 + [y - (1/2)]^2

1/2 = [x - (1/2)]^2 + [y - (1/2)]^2

This is a circle centered at (1/2, 1/2), with radius sqrt(1/2). Graph the circle as a dotted line.

Now, going back to the inequality

x + y > x^2 + y^2

If we test the point (1/2,1/2), we will get

1/2 + 1/2 > (1/2)^2 + (1/2)^2 , or
1 > 1/2, which is a true statement. Therefore, shade the inside of the circle.

There's another restriction; you cannot take the log of a negative number, so

x + y > 0
Therefore
y > -x, so graph y = -x as a dotted line (which should cut the circle into two).
If we test the point (1/2, 1/2) for this inequality, we get

1/2 > -1/2, which is a true statement. Shade the side of the line which contains the point (1/2, 1/2), and you'll now have your figure.

Answer 2

For log to be greater than 1, both base and number should be either between 0 & 1 or greater than 1 i.e

0< x+y < 1 & 0 < x^2 + y^2 <1
Or
x+y >1 & x^2 + y^2 > 1