<s>[INST] 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 $0 < x < 2x_0$)?
 [/INST] Hint: $\log x$ is concave, so it remains below its tangent at $x_0$.
</s>