In slope-intercept form, what is the equation of the line tangent to y = ln x at the point (e, 1)?

In slope-intercept form, what is the equation of the line tangent to y = ln x at the point (e, 1)?

• A. y = (1/e)x
• B. y = ex
• C. y = (1/e)x + 1
• D. y = ln(x) + 1

Answer: (A)

To find the equation of the line tangent to the curve defined by \( y = \ln x \) at the point \( (e, 1) \), we first need to determine the slope of the tangent line at that point. The slope of the curve at any point can be found by taking the derivative of the function. The derivative of \( y = \ln x \) is given by \( y' = \frac{1}{x} \). To find the slope at the point \( (e, 1) \), we substitute \( x = e \) into the derivative: \[ y'(e) = \frac{1}{e}. \] This means the slope of the tangent line at the point \( (e, 1) \) is \( \frac{1}{e} \). Next, we use the point-slope form of the equation of a line, which is \( y - y_0 = m(x - x_0) \), where \( (x_0, y_0) \) is the point on the curve, and \( m \) is the slope. Here, \( (x_0, y_0) = (e, 1

To find the equation of the line tangent to the curve defined by ( y = \ln x ) at the point ( (e, 1) ), we first need to determine the slope of the tangent line at that point. The slope of the curve at any point can be found by taking the derivative of the function.

The derivative of ( y = \ln x ) is given by ( y' = \frac{1}{x} ). To find the slope at the point ( (e, 1) ), we substitute ( x = e ) into the derivative:

[

y'(e) = \frac{1}{e}.

]

This means the slope of the tangent line at the point ( (e, 1) ) is ( \frac{1}{e} ).

Next, we use the point-slope form of the equation of a line, which is ( y - y_0 = m(x - x_0) ), where ( (x_0, y_0) ) is the point on the curve, and ( m ) is the slope. Here, ( (x_0, y_0) = (e, 1