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!

1 Answer
Mar 21, 2018

y=x/2-1+ln(2)y=x21+ln(2), or, written only in terms of logs, y=x/2+ln(2/e).y=x2+ln(2e).

Explanation:

First, let's differentiate y=ln(x).y=ln(x). The derivative of a function in terms of xx gives us the rate of change (or slope) of that function at some value for x.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)