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

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.

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.

Okay, thank you very much.

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

3/e = 1.1036

To solve part A, we need to find the tangent line approximation for ln 3.

The point-slope form of a linear equation is given by y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.

In this case, the slope can be found by differentiating the function ln(x) with respect to x. Given that d(ln x)/dx = 1/x, when x > 0, we can substitute x = 3 into the derivative equation to find the slope:

m = d(ln x)/dx = 1/x = 1/3.

Now we need to determine a point on the line. We are told that lne = 1, so (e, 1) is a point on the graph of ln(x).

Using the point-slope form, we have y - 1 = (1/3)(x - e).

To simplify further and find the equation of the tangent line, we can distribute 1/3 to (x - e):

y - 1 = (1/3)x - e/3.

To isolate y, we can add 1 to both sides:

y = (1/3)x - e/3 + 1.

Now we have the equation of the tangent line approximation for ln 3.

Regarding part B, to calculate the percent error in the approximation, we typically use the formula:

% error = (|exact value - approximate value| / |exact value|) * 100.

In this case, the exact value is ln 3, and the approximate value is the y-coordinate of the tangent line at x = 3.

To find the exact value of ln 3, we can use a calculator or table of logarithms. ln 3 is approximately 1.0986.

To find the approximate value, we substitute x = 3 into the equation of the tangent line:

y = (1/3)(3) - e/3 + 1 = 1 - e/3 + 1 = 2 - e/3.

So the approximate value is 2 - e/3.

Using these values, we can calculate the percent error:

% error = (|1.0986 - (2 - e/3)| / |1.0986|) * 100.

Simplifying further:

% error = (|1.0986 - 2 + e/3| / 1.0986) * 100
= (|0.9014 + e/3| / 1.0986) * 100.

Since e is Euler's number (approximately equal to 2.71828), we can substitute it:

% error = (|0.9014 + 2.71828/3| / 1.0986) * 100
= (|0.9014 + 0.906093| / 1.0986) * 100
≈ (|1.807493| / 1.0986) * 100
≈ (1.807493 / 1.0986) * 100
≈ 164.44%.

Therefore, the percent error in the approximation is approximately 164.44%.