Given that X1 and X2 are the two roots of the equations x + log3x = - 1 and X + 3 * x = - 1 respectively, how much is X1 + x2 equal to

Given that X1 and X2 are the two roots of the equations x + log3x = - 1 and X + 3 * x = - 1 respectively, how much is X1 + x2 equal to

Solution 1
-1-x=3^x
-1-x=log3x
F (x) = 3 ^ X and G (x) = log3x are reciprocal functions
So their images are symmetric with respect to y = X
The image of - 1-x is also symmetric with respect to y = X
Draw an image on the coordinate system
The intersection of these two functions with - 1-x is symmetric with respect to y = X
So suppose the coordinates of the two intersections are (x1, Y1) (X2, Y2)
The slope is (y1-y2) / (x1-x2) = - 1
And the midpoint has (x1 + x2) / 2 = (Y1 + Y2) / 2 on the line y = X
The solution is X1 = Y2, Y1 = x2
So - 1-x1 = 3 ^ x1
3^x1=y1=x2
-1-x1=x2
x1+x2=-1
Solution 2
x1+log(3)x1=-1
3^(-1-x1)=x1
3^(-1-x1)=-1-(-1-x1)
-1-x1+3^(-1-x1)=-1
It is found that - 1-x1 is the solution of the equation x + 3 ^ x = - 1
Because both functions are incremental, there is only one real solution
So - 1-x1 = X2, that is, X1 + x2 = - 1

If X1 is the solution of the equation log3x = 0 and X2 is the solution of 9 ^ x = 2.3 ^ x + 3, then X1 + x2 =?

Log3 x = 0, X1 = 1; 3 ^ (2x) = 2 × 3 ^ x + 3 (3 ^ x) ^ 2-2 (3 ^ x) - 3 = 0 (3 ^ x + 1) (3 ^ x-3) = 0, but 3 ^ x + 1 > 0, so only 3 ^ x-3 = 0, X2 = 1. So X1 + x2 = 1 + 1 = 2