Let [tex]f(x,y,z)=z-\ln(x+y)[/tex]. The gradient of [tex]f[/tex] at the point (1, 0, 0) is the normal vector to the surface, which is also orthogonal to the tangent plane at this point.
So the tangent plane has equation
[tex]\nabla f(1,0,0)\cdot(x-1,y,z)=0[/tex]
Compute the gradient:
[tex]\nabla f(x,y,z)=\left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z}\right)=\left(-\dfrac1{x+y},-\dfrac1{x+y},1\right)[/tex]
Evaluate the gradient at the given point:
[tex]\nabla f(1,0,0)=(-1,-1,1)[/tex]
Then the equation of the tangent plane is
[tex](-1,-1,1)\cdot(x-1,y,z)=0\implies-(x-1)-y+z=0\implies\boxed{z=x+y-1}[/tex]
