To find the minimum value of \( \frac{x}{\log x} \), we can use calculus. This particular expression is an optimization problem, which is suitable for finding maxima or minima using differentiation.
\(f(x) = \frac{x}{\log x}\) where \(\log x\) is the natural logarithm, denoted typically as \(\ln x\).
Let us denote \(y = \frac{x}{\ln x}\). To find the minimum, we first find the derivative \(\frac{dy}{dx}\) using the quotient rule:
If \(u = x\) and \(v = \ln x\), then
Derivative of \(u\) (\(x\)) is 1.
Derivative of \(v\) (\(\ln x\)) is \(\frac{1}{x}\).
Using the quotient rule:
Applying this, we get: \[ \frac{dy}{dx} = \frac{\ln x \cdot (1) - x \cdot \left(\frac{1}{x}\right)}{(\ln x)^2} = \frac{\ln x - 1}{(\ln x)^2} \]
Set \(\frac{dy}{dx} = 0\) to find critical points: \[ \frac{\ln x - 1}{(\ln x)^2} = 0 \implies \ln x - 1 = 0 \implies \ln x = 1 \] Solving \(\ln x = 1\) gives: \[ x = e \]
To confirm this is a minimum, examine the second derivative \( \frac{d^2y}{dx^2} \). If it is positive at \( x = e\), then \( x = e \) is indeed a minimum.
Upon calculating the second derivative and evaluating, it turns out \( \frac{d^2y}{dx^2} \) is positive at \( x = e \), confirming a minimum.
Therefore, the minimum value of \( \frac{x}{\ln x} \) is when \( x = e \), and the minimum value is \( e \).