Find the slope of the normal line to y= ln(15-x) at x=4
when x=4, y = ln(11)
so our point is (4,ln11)
dy/dx = -1/(15-x)
which, when x=4 is -1/11
so the slope of the normal is +11
equation:
y - ln11 = 11(x-4)
11x - 44 = y-ln11
11x - y = 44 - ln11 OR y = 11x + ln11 - 44