How to compare the values $\sqrt 2$ and $\ln(3).$ How to compare the values $\sqrt 2$ and $\ln(3)?$ I know only $\ln(x)<x$ and $\sqrt{x}<x$. Please help. Thanks.

The trick is to look at $f(x) = \sqrt{x} - \ln(1+x)$.
Note that for $x = 1$, 
$$1 - \ln(2) > 0$$
This is true since $\ln(x)$ is an increasing function and $1 < 2 < e$ implies
$$0 = \ln(1) < \ln(2) < \ln(e) = 1$$
Now look at the derivative
$$f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{x+1}$$
We would like to show that $f'(x) \ge 0$ for $x > 1$, which means $f(x)$ is increasing and must therefore be positive for $x > 1$ (since it is already positive at $x=1$ per above).
Consider the following equivalent inequalities (when $x > 1$):
$$\begin{eqnarray}
\frac{1}{2\sqrt{x}} - \frac{1}{x+1} &\ge& 0\\
\frac{1}{2\sqrt{x}} &\ge& \frac{1}{x+1}\\ 
x + 1 &\ge& 2\sqrt{x}                                   
\end{eqnarray}$$
For the last inequality, $1+1 = 2 = 2\sqrt{1}$ and for $x > 1$, taking the derivatives agains shows that the left side grows at a rate of $1$ and the right side grows at the rate of $\frac{2}{2\sqrt{x}} = \frac{1}{\sqrt{x}} < 1$ when $x > 1$.
Thus the last equivalent inequality holds for $x > 1$, which implies $f'(x) \ge 0$ for $x > 1$ which implies $f(x)$ is positive for $x > 1$. In particular, $f(2) = \sqrt{2} - \ln(3) > 0$. Therefore $\sqrt{2} > \ln(3)$.
