y = log(x). how do you find the tangent line at x = 2?

Answer 1

First, you need the slope of the function y = log(x). That is, you need y'. Then, plug in x = 2 into your y' equation. This will be the slope of the line you are after. Once that is done you need to determine y's value when x = 2 which is obviously log(2). At this point (no pun intended) you have the slope and a point, (2, log(2)) on the line you are looking for. Use either the point-slope formula for a straight line or plug in your point and slope into the equation y = mx + b and determine b. At this point you are done with the problem.

This problem is simply asking you to find a linear approximation of the function y = log(x) in the neighborhood around x = 2.

Repost with your answer.

Update: Here is a quick video solution for your problem.
http://screencast.com/t/UDSFTpa2

Answer 2

You will need to find the slope of y = log(x).
The equation for the slope is the derivative. I will assume the log(x) is not the natural of of x and since the base is not given the base is 10.
y' = log base 10(e)*1/x which I will write y'=log(e)/x
the slope at 2 is m = log(e)/2 = .217147.
A point on the line is needed and is found by substituting into the original function. (2, log (2))
Y-log(2) = log(e)/2*(x-2)
Y = log(e)/2*x + log(e)+log(2)
In decimal form:
Y = .217147x +.7353
Check the question again to make sure it was not the natural log(x) or ln(x)

Answer 3

y = log(x)
y' = 1/x = 1/2 at at x = 2
y= mx +b
y = x/2 +b
When x = 2 ,y = log(2)
log(2)= 2/2 + b
b= log(2)-1
So y = x/2 +log(2) -1