So the problem is "For x>0, d(lnx)/dx = 1/x and lne=1.

Question

A) find the tangent line approximation for ln 3
B) calculate the %error in approximation in part A.

My teacher said the answer for part A is that 1/e, and I'm pretty confused. So far I the given info into point slope form of y-1=1/3(x-e) and I in stuck on how to solve from there.

Answer 1

clearly 1/e is no kind of approximation for ln3.

However, if we use the tangent at (e,1) (with slope 1/e), the line is

y-1 = (1/e)(x-e)

Then extending the line to x=3,
y-1 = 1/e (3-e)
y = 3/e

So, 3/e is an approximation to ln3.

4-e = 1.1036
ln3 = 1.0986

The estimate is high, since the curve is concave down.

Answer 2

Okay, thank you very much.

Answer 3

sorry about the typo. 4-e was from a previous incorrect solution. It should have read

3/e = 1.1036