Answer
The tangent line to the curve is parallel to $3x-y=5$ on $(\ln 3,7-3\ln 3)$.
Work Step by Step
Given a curve $y=1+2e^x-3x$ and a line $l:3x-y=5$
Find the slope of $l$:
$3x-y=5$
$y=3x-5$
$m_l=3$
Find $\frac{dy}{dx}$:
$\frac{dy}{dx}=\frac{d}{dx}(1+2e^x-3x)=0+2e^x-3=2e^x-3$
Find $x$ such that the tangent line to the curve is parallel to $l$, that is $\frac{dy}{dx}=m_l$:
$2e^x-3=3$
$2e^x=6$
$e^x=3$
$x=\ln 3$
Find $y$ when $x=\ln 3$:
$y=1+2e^{\ln 3}-3\ln 3=1+2\cdot 3-3\ln 3=7-3\ln 3$
Find the equation of the tangent line to the curve at $(\ln 3,7-3\ln 3)$:
$y-(7-3\ln 3)=3(x-\ln 3)$
$y-7+3\ln 3=3x-3\ln 3$
$y=3x+7-6\ln 3$
Graph the curve, the tangent line, and the line $l$:
