Derive an upper bound for $\log(x)$

Question

A paper has given that for $x_0 > 0$,

$$ \log x \leq \frac{x}{x_0} + \log x_0 - 1. $$

The way I tried to prove this is:

For $|t| < 1$, \begin{align*} \log(1 - t) = -\sum_{k=1}^n \frac{t^k}{k} \end{align*} We can replace $t = 1 - \frac{x}{x_0}$, \begin{align*} \log x - \log x_0 & = - [(1 - \frac{x}{x_0}) + \frac{1}{2}(1 - \frac{x}{x_0})^2 + ...]\\ & = -1 + \frac{x}{x_0} - [\frac{1}{2}(1 - \frac{x}{x_0})^2 + \frac{1}{3}(1 - \frac{x}{x_0})^3 + ...]\\ & = -1 + \frac{x}{x_0} - c \end{align*}

How can we show that $c$ is positive and what's the range of $x$ (is it $0 < x < 2x_0$)?

Answer 1

For any real $w$, by the Taylor formula with Lagrange remainder, we have $$ e^{w - 1} = 1 + (w - 1) + \frac{{(w - 1)^2 }}{2}e^\xi \ge 1 + (w - 1) = w $$ with some $\xi$ between $0$ and $w-1$. Thus, for any $w>0$, $w-1 \ge \log w$. With $w=x/x_0$, where $x$, $x_0>0$, we get $$ \log \left( {\frac{x}{{x_0 }}} \right) \le \frac{x}{{x_0 }} - 1 \Longleftrightarrow \log x \le \frac{x}{{x_0 }} + \log x_0 - 1. $$

Answer 2

Let $a:=x/x_0 >0$, where $x, x_0 >0$.

Want to show $\log a \le a-1.$

Recall

$\displaystyle{\int_{1}^{a}}(1/t)dt =\log a;$

1)$a\ge 1:$

$\log a = \displaystyle{\int_{1}^{a}}(1/t)dt \le \displaystyle{\int_{1}^{a}}(1)dt,$

$\log a \le a-1.$

2)$1>a>0:$

Then $\displaystyle{\int_{a}^{1}}(1/t)dt >\displaystyle{\int_{a}^{1}}(1)dt=1-a;$

$(-1)\displaystyle{\int_{a}^{1}} (1/t)dt <a-1;$

$\log a= \displaystyle{\int_{1}^{a}}(1/t)dt <a-1$, and we are done.

Answer 3

Hint: $\log x$ is concave, so it remains below its tangent at $x_0$.