Calculus (3rd Edition)

$$y=16((\ln \sqrt{2} ) \ x+1-4\ln 2).$$
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 $$y=16((\ln \sqrt{2} ) \ x+1-4\ln 2).$$