Tangents & Normals to Logarithmic Curves
Tangents and normals to logarithmic curves use \(\dfrac{d}{dx}\ln x=\dfrac1x\) to find the gradient, then form the tangent \(y-y_0=m(x-x_0)\) and its perpendicular normal.
Part of the NSW Year 12 Mathematics Advanced course, in the Calculus area of study (Differential calculus focus area) of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.
Theory
Tangents to a log curve use the gradient \(\dfrac{d}{dx}\ln x=\dfrac1x\). This Year 12 Mathematics Advanced topic (MAV-12-04) finds tangents and normals to logarithmic curves.
At \((x_0,y_0)\) on \(y=f(x)\): the tangent has gradient \(m=f'(x_0)\), equation \(y-y_0=m(x-x_0)\); the normal has gradient \(-\dfrac1m\).
For \(y=\ln x\) the gradient at \(x_0\) is \(\dfrac{1}{x_0}\), and the curve passes through \((1,0)\).
Method
- Find the point \((x_0,\ln x_0)\).
- Find the gradient \(m=\dfrac{1}{x_0}\).
- Write the tangent (or normal with \(-\tfrac1m\)).
| \(m\) | \(=\) | \(\dfrac11=1\) |
| \(y\) | \(=\) | \(x-1\) |
| \(\text{point}\) | \((e,1)\) | |
| \(m\) | \(=\) | \(\dfrac1e\) |
| \(y\) | \(=\) | \(\dfrac{x}{e}\) |
| \(\text{gradient}\) | \(=\) | \(-1\) |
| \(y\) | \(=\) | \(-x+1\) |
| \(m\) | \(=\) | \(\dfrac12\) |
Common pitfalls
Frequently asked questions
How do you find the tangent to y = ln x?
Find the point, compute the gradient as 1 over the x-value, then write y minus y nought equals m times (x minus x nought).
What is the gradient of y = ln x at x = 1?
It is 1 over 1, which is 1, so the tangent there is y equals x minus 1.
Where does y = ln x cross the x-axis?
At the point (1, 0), because ln 1 equals 0.
What is the normal to a logarithmic curve?
It is the line perpendicular to the tangent at the point, with gradient minus 1 over the tangent gradient.