Question

What is the equation of the normal line of `f(x)=ln2x-x` at `x=4`?

Answer 1

`y=4/3x+3ln2-28/3`

Explanation:

The equation of the line normal to the curve `y=f(x)` in the point `x=bar x` is given by:

`y=f(bar x) -1/(f'(bar x))(x-barx)`

Given `bar x = 4`:

`f(x) = ln2x-x`
`f(4) =ln8-4=3ln2-4`

`f'(x) = 1/x-1`
`f'(4) = -3/4`

The equation of the normal line is:

`y=4/3(x-4)+3ln2-4=4/3x+3ln2-28/3`