Calculus of a Single Variable, Chapter 5, 5.5, Section 5.5, Problem 61

y=log_3(x)
The line is tangent to the graph of the function at (27,3). The equation of the tangent line is _____.
To solve, the slope of the tangent line should be determined. Take note that the slope of a tangent line is equal to the value of the derivative at the point of tangency.
To determine the derivative of the function, apply the formula d/dx[log_a(u)]=1/(ln(a)*u)*(du)/dx.
(dy)/dx = d/dx[log_3 (x)]
(dy)/dx =1/(ln(3)*x) * d/(dx)(x)
(dy)/dx =1/(ln(3)*x)*1
(dy)/dx = 1/(xln(3))
The point of tangency is (27,3). So plug-in x = 27 to the derivative to get the slope of the tangent.
m=(dy)/dx = 1/(27ln(3))
Hence, the line that is tangent to the graph of the function at point (27,3) has a slope of m = 1/(27ln(3)) .
To get the equation of the line, apply the point-slope form.
y-y_1=m(x - x_1)
Plugging in the values of m, x1 and y1, it becomes:
y - 3= 1/(27ln(3))(x - 27)
y-3=1/(27ln(3))*x - 27 * 1/(27ln(3))
y - 3 =x/(27ln(3)) - 1/ln(3)
y =x/(27ln(3)) - 1/ln(3)+3

Therefore, the equation of the tangent line is y =x/(27ln(3)) - 1/ln(3)+3 .