Answer
$$ y=16((\ln \sqrt{2} ) \ x+1-4\ln 2).$$
Work Step by Step
Let $ f(x)=\sqrt{2}^x $, then $ f'(x)=\sqrt{2}^x \ln \sqrt{2}$ and the slope of $ f $ at $ x=8$ is given by $$ m=f'(8)=\sqrt{2}^8 \ln \sqrt{2}.=16 \ln \sqrt{2}$$ The equation of the tangent line $$ y=16(\ln \sqrt{2} ) \ x+c.$$ Since the function and the tangent line coincide at $ x=8$, we have
$$ c=\sqrt{2}^8-16(\ln \sqrt{2} )(8)=16(1-4\ln 2).$$ Finally, we get $$ y=16((\ln \sqrt{2} ) \ x+1-4\ln 2).$$
