Question

Equation of the tangent to curve using logs?

Find the equation of the tangent to the curve y= ln(x) at the point (2, ln(2))
I just can't seem to work this one out! Any help is appreciated!

Answer 1

`y=x/2-1+ln(2)`, or, written only in terms of logs, `y=x/2+ln(2/e).`

Explanation:

First, let's differentiate `y=ln(x).` The derivative of a function in terms of `x` gives us the rate of change (or slope) of that function at some value for `x.` Thus, we can use the derivative to find the slope of our tangent line at the given coordinates.

`y'=1/x` (The derivative of the natural logarithm function is `1/x`).

We want the slope at `(x_0,y_0)=(2,ln(2)),` so we evaluate our derivative at `x=2.` Note that the value for `y` plays absolutely no role here:

`y'(2)=1/2`

Now, we use the point-slope form of the line to determine the tangent line equation:

`y-y_0=m(x-x_0)` where `m` is our slope and `(x_0,y_0)` is some point on the line.

`m=1/2,` as calculated above.

`(x_0, y_0)=(2,ln(2))` as given to us by the problem. Thus, our tangent line equation becomes

`y-ln(2)=1/2(x-2)`

Solve for `y:`

`y-ln(2)=x/2-1`

`y=x/2-1+ln(2)`

We could further write `1` as `ln(e)` and continue simplifying. Totally optional.

`y=x/2-ln(e)+ln(2)`

`y=x/2+ln(2)-ln(e)`

`y=x/2+ln(2/e)` as `ln(a)-ln(b)=ln(a/b)`