Exercise 1270Calculus › Single-variable calculus › Differentiation › Mean Value TheoremL1Added 4 Nov 2026

A logarithm inequality via the MVT

#mean value theorem#inequality#logarithm

Statement

Prove that for all $x>1$ : $\frac{x-1}{x}<\ln x<x-1$ .

Hint

Apply the MVT to $\ln$ on $[1,x]$ ; bound $\frac{1}{c}$ between $1/x$ and $1$ .

Show solutionHide solution

MVT. There exists $c\in]1,x[$ with $\ln x-\ln1=\frac{1}{c}(x-1)$ .

Bounds. Since $1<c<x$ , $\frac{1}{x}<\frac{1}{c}<1$ . Multiply by $x-1>0$ :
$\frac{x-1}{x}<\ln x<x-1.$

Conclusion. The double inequality holds.