📂Lemmas

Inequalities for the Logarithmic Function 1-1/x < log x < x-1

Theorem

For a logarithmic function with base $e$, the following inequality holds:

$$1 - \dfrac{1}{x} \le \ln x \le x - 1\qquad \text{ for } x \gt 0$$

Proof¹

Part 1. $\ln x \le x - 1$

Let’s set it as $f(x) = x - 1 - \ln x$. Differentiating it gives, $f^{\prime}(x) = 1 - \dfrac{1}{x}$ $(x>0)$.

• At $0 \lt x \lt 1$, it is $f^{\prime} \lt 0$
• If $x = 1$, then $f^{\prime} = 0$
• At $x \gt 1$, it is $f^{\prime} \gt 0$

Since $f^{\prime}(1) = 0$, $f$ has a minimum value of $0$ at $1$. Therefore,

$$0 \le f(x) \implies 0 \le x - 1 - \ln x \implies \ln x \le x - 1 \qquad \text{ for } x > 0$$

Part 2. $1 - \dfrac{1}{x} \le \ln x$

Let’s set it again as $f(x) = \ln x - 1 + \dfrac{1}{x}$. Differentiating it gives, $f^{\prime}(x) = \dfrac{1}{x} - \dfrac{1}{x^{2}} = \dfrac{1}{x}\left( 1 - \dfrac{1}{x} \right)$ $(x > 0)$.

• At $0 \lt x \lt 1$, it is $f^{\prime} \lt 0$
• If $x = 1$, then $f^{\prime} = 0$
• At $x \gt 1$, it is $f^{\prime} \gt 0$

Since $f^{\prime}(1) = 0$, $f$ has a minimum value of $0$ at $1$. Therefore,

$$0 \le f(x) \implies 0 \le \ln x - 1 + \dfrac{1}{x} \implies 1 - \dfrac{1}{x} \le \ln x \qquad \text{ for } x > 0$$

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