Sketching logarithmic curves

Tangents and normals to logarithmic curves, and finding stationary points and sketching curves involving ln x, paying attention to the domain.

Tangents and normals

Example 1●●

Find the tangent and normal to $y = \ln x$ at $x = e$.

The point is $(e, 1)$ and the gradient is $\dfrac1e$.

Point and gradient

$y - 1 = \dfrac1e(x - e)$, so $y = \dfrac{x}{e}$. It passes through the origin.

Tangent

The gradient is $-e$: $y - 1 = -e(x - e)$, so $y = -ex + e^2 + 1$.

Normal

Example 2: y = x ln x●●●

Sketch $y = x\ln x$.

The domain is $x > 0$. $y = 0$ when $\ln x = 0$, at $x = 1$.

Domain and intercept

$y' = \ln x + 1 = 0$ gives $x = \dfrac1e$, with $y = -\dfrac1e$.

Stationary point

$y'' = \dfrac1x > 0$, so it is a minimum, and the curve is concave up everywhere.

Nature

As $x \to 0^+$, $y \to 0$: the curve heads to the origin, which is not included.

Near zero

y = x ln x

The minimum is at x = 1/e ≈ 0.37, where the gradient is zero.

Example 3●●

Find and classify the stationary point of $y = \dfrac{\ln x}{x}$.

$y' = \dfrac{1 - \ln x}{x^2} = 0$ gives $\ln x = 1$, so $x = e$ and $y = \dfrac1e$.

For $x < e$, $y' > 0$, and for $x > e$, $y' < 0$. So $\left(e, \tfrac1e\right)$ is a maximum.

Flashcards

Domain of $ln f(x)$

$f(x) > 0$

Minimum of $y = xln x$

$\left(\tfrac1e, -\tfrac1e\right)$

Maximum of $y = dfrac{ln x}{x}$

$\left(e, \tfrac1e\right)$

Practice