Question

At what point on the curve `y=1+2e^x-3x` is the tangent line parallel to the line `3x-y=5`?

Answer 1

`(ln3,7-3ln3)`

Explanation:

`3x-y=5`
`y=3x+5`

Here we have slope `m=3` since it is in the form `y=mx+b`.
Since the lines are parallel, they have the same slope.

Find the derivative to find the point wih tangent line of slope `3`:
`y=1+2e^x-3x`
`y'=0+2e^x-3`
`3=2e^x-3`
`3=e^x`
`x=ln3`

Find the `y`-coordinate.
`y=1+2e^x-3x`
`y=1+2e^(ln3)-3ln3`
`y=7-3ln3`

Thus it is parallel to the tangent line at the point `(ln3,7-3ln3)`.