Use the tangent line approximation to approximate the value of ln(1008)

assuming you want to use an "easily" calculatable value, let's approximate assuming we know the value of ln(1000).

the slope of the tangent at any point (x,y) on the curve is

m = 1/x

so, at x=1000, the slope is 1/1000.

Thus, we are using the line through (1000,ln1000) with slope 1/1000:

y = 1/1000 (x-1000) + ln1000
at x=1008,
y = 8/1000 + ln1000 = 6.915755

real value: 6.915723

To approximate the value of ln(1008) using the tangent line approximation, we can start by finding the tangent line to the natural logarithm function at a point we know well. In this case, let's choose the point (1000, ln(1000)).

First, we need to find the derivative of the natural logarithm function (ln(x)). The derivative of ln(x) is 1/x. Next, we need to find the slope of the tangent line at (1000, ln(1000)) by evaluating the derivative at x=1000. The slope is thus 1/1000.

Now that we have the slope, we can form the equation of the tangent line using the point-slope form:

y - ln(1000) = (1/1000)(x - 1000).

Next, we need to plug in the value of x=1008 into the equation of the tangent line to approximate the value of ln(1008).

y - ln(1000) = (1/1000)(1008 - 1000).

Simplifying the equation gives us:

y - ln(1000) ≈ (1/1000)(8).

Now, we can solve for y to find the approximate value of ln(1008):

y = ln(1000) + (1/1000)(8).

Calculating this result will give you the approximate value of ln(1008) using the tangent line approximation.