Find the equation of the line tangent to the graph of f(x)= (ln x)^3 at x=11 The answer is in slope intercept form Thank you!?

DWRead2015-04-22T10:22:48Z

f(11) = ln³(11) ≅ 13.79
f'(x) = 3ln²(x)/x
f'(11) = 3ln²(11)/11 ≅ 1.57

y-13.79 = 1.57(x-11)

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