Answer
$$y=(\ln2 )x+1$$
Work Step by Step
Re-arranging the function into a differentiable form:
$y=e^{(\ln2) x}$
The derivative of the curve is (using the chain rule):
$y'=\frac{de^{(\ln2)x}}{d(\ln2)x}
\times\frac{d(\ln2)x}{dx}$
$=e^{(\ln 2)x} \times \ln2$
$=\ln 2\times2^x$
Thus, the slope ($m$) of the tangent line at (0,1) would be:
$m= y'(0)$
$=\ln 2 \times 2^{(0)}$ (plugging $x=0$ into equation $y'$)
$=\ln2$
Thus, plugging in the given points (0,1) into the general point-slope form of a linear equation, an equation of the tangent line is:
$y-1=\ln2(x-0)$
Further simplifying the equation gives:
$y=(\ln2 )x+1$
