Calculus of a Single Variable, Chapter 5, 5.5, Section 5.5, Problem 62

y = log_10 (2x)
The line is tangent to the graph of function at (5,1). The equation of the tangent line is _______.
To solve this, we have to determine the slope of the tangent. Take note that the slope of a tangent is equal to the derivative of the function.
To get the derivative of the function, apply the formula d/dx [log_a (u)] = 1/(ln(a)*u)*(du)/dx .
Applying this, the y' will be:
y' = d/dx [log_10 (2x)]
y' = 1/(ln(10) * 2x) * d/dx (2x)
y'=1/(ln(10) * 2x) *2
y'=1/(xln(10))
Then, plug-in the given point of tangency to get the slope.
y'= 1/(5ln(10))
So the line that is tangent to the graph of the function at point (5,1) has a slope of m=1/(5ln(10)) .
To get the equation of the line, apply the point-slope form
y - y_1 = m(x - x_1)
Plugging in the values, it becomes:
y - 1 = 1/(5ln(10))(x - 5)
y - 1 = 1/(5ln(10))*x - 1/(5ln(10))*5
y - 1 =x/(5ln(10)) -1/(ln(10))
y=x/(5ln(10)) - 1/(ln(10))+1

Therefore, the equation of the tangent line is y=x/(5ln(10)) - 1/(ln(10))+1 .