posted by paul .
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
y = 8/1000 + ln1000 = 6.915755
real value: 6.915723