Answer
$z=x-2y-1$
OR
$x-2y-z-1=0$
Work Step by Step
$z=ln(x-2y)$, $(3,1,0)$
Consider $f(x,y)= ln(x-2y)$
$f_{x}(x,y)=\frac{1}{x-2y}$
$f_{y}(x,y)=\frac{-2}{x-2y}$
At $(3,1,0)$
$f_{x}(3,1)=\frac{1}{3-2.1}=1$
$f_{y}(3,1)=\frac{-2}{3-2.1}=-2$
The equation of the tangent plane to the given surface at the specified point $(3,1,0)$is given by
$z-z_{0}=f_{x}(x_{0},y_{0})(x-x_{0})+f_{y}(x_{0},y_{0})(y-y_{0})$
On substituting the values, we get
$z-0=1(x-3)+(-2)(y-1)$
$z=x-3-2y+2$
Hence, $z=x-2y-1$ OR $x-2y-z-1=0$