Derive an upper bound for $\log(x)$ 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 $0 < x < 2x_0$)?

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
$$
\log \left( {\frac{x}{{x_0 }}} \right) \le \frac{x}{{x_0 }} - 1 \Longleftrightarrow \log x \le \frac{x}{{x_0 }} + \log x_0  - 1.
$$
