#### Answer

$y=3x-3(\ln3-1)$

#### Work Step by Step

$y=e^{x}$ $;$ $a=\ln3$
First, evaluate the derivative of the given expression:
$y'=e^{x}$
Substitute $x$ by $a=\ln3$ in the derivative found to obtain the slope of the tangent line at the given point:
$m_{tan}=e^{\ln3}=3$
Substitute $x$ by $a=\ln3$ in the original expression to obtain the $y$-coordinate of the point given:
$y=e^{\ln3}=3$
The point is $(\ln3, 3)$
The slope of the tangent line and a point through which it passes are now known. Use the point-slope form of the equation of a line, which is $y-y_{1}=m(x-x_{1})$ to obtain the equation of the tangent line at the given point:
$y-3=3(x-\ln3)$
$y-3=3x-3\ln3$
$y=3x-3\ln3+3$
$y=3x-3(\ln3-1)$
The graph of both the function and the tangent line are shown in the answer section.