Answer
The line parallel to $3x-y=5$ that is tangent to the curve is $y=3x-6 \ln3 +7$ at the point $(1.1, 3.7)$ or exactly $(\ln3, 7-3\ln3)$.
Work Step by Step
We can find the equation of the tangent line by first finding the derivative of the curve.
This equation is $y=2e^x-3x$.
Plugging in $3x-y=5$'s slope of $3$ for $y$ in this new equation gives us the x-coordinate at which the original curve's slope is $3$.
This x-coordinate is $\ln3$.
We can now plug this x-coordinate into the original curve's equation to find the y-coordinate of $7-3\ln3$.
So the point where the line parallel to $3x-y=5$ is tangent to the curve is $(\ln3, 7-3\ln3)$.
