How do I square a logarithm? How do I square $\log_2(3)$. Does it become $2\log_2(3)$ ?
 A: $\log_2(3) \approx 1.58496$ as you can easily verify.
$(\log_2(3))^2 \approx (1.58496)^2 \approx 2.51211$.
$2 \log_2(3) \approx 2 \cdot 1.58496 \approx 3.16992$.
$2^{\log_2(3)} = 3$.
Do any of those appear to be equal?
(Whenever you are wondering whether some general algebraic relationship holds, it's a good idea to first try some simple numerical examples to see if it is even possible.)
Actually, the only way that $(\log_2(3))^2 = 2 \log_2(3)$ could hold is if $\log_2(3)$ were equal to 2 or 0.  That is clearly false since $2^2 = 4 \ne 3$ and $2^0 = 1 \ne 3$.
A: No.The best you can do is $$(\log_2 3)^2=\log_2 3\cdot \log_2 3=\log_2 (3^{\log_2 3})$$
A: No. $(\log_2(3))^2$ can't be simplified.
However, $\log_2(3^2)=2\log_2(3)$.
A: Agreeing with the rest of the answers here, you cannot simplify any further.  You could, however, do a change of base with the logs and put them in base $10$.  We have the formula
$$\log_bx=\frac{\log_ax}{\log_ab}$$where $a$ can be any base you want.  Most common base is $10$.  So we have,
$$(\log_23)^2=\left(\frac{\log_{10}3}{\log_{10}2}\right)^2=\left(\frac{\log3}{\log2}\right)^2=\frac{\log^23}{\log^22}$$
A: No, it doesn't. Logarithms follow this rule:
$$
\log_b (a^c) = c\log_b a,
$$
while your statement says that
$$
(\log_b a)^c=c\log_b a,$$
which is basically saying the same as $x^y=yx$.
A: \begin{eqnarray}
(\log_{2}(3))^{2} &=& (\log_{2}(3))(\log_{2}(3)) \\
                  &=& \log_{2}(3^{\log_{2}(3)})
\end{eqnarray}
That's as simple as it gets!
