Comparing numbers in form $x^y$ Let's consider two numbers in form $x_1^{y_1}$ and $x_2^{y_2}$ 
How can we compare those two numbers without evaluating them ? 
Can we use logarithms to check it ? If yes - how ?
Thanks in advance.
P.S It's not my homework :) 
 A: Yes, we can. The key is that 
$$\log_b a = \frac{\log_{b'} a}{\log_{b'} b}$$
and that taking logarithms of positive numbers (I assume you're working with positive $x_1,x_2$) with respect to the same base preserves inequalities. Thus we can take $\log_{x_1} x_1^{y_1} = y_1$
and 
$$\log_{x_1} x_2^{y_2} = \frac{\log_{x_2} x_2^{y_2}}{\log_{x_2} x_1} = \frac{y_2}{\log_{x_2}{x_1}}$$
and compare these two numbers instead, which is easier.
A: Let's assume $x_i > 1$ and $y_i > 0$, because other cases can be handled analogously (just be careful when negatives crop up).  Certain cases are easy.  If both $x_1 \geq x_2$ and $y_1 \geq y_2$, then $x_1^{y_1} \geq x_2^{y_2}$.  The more interesting cases occur when there is a mixed relationship, such as $x_1 \leq x_2$ and $y_1 \geq y_2$. In fact, let's assume the latter two inequalities and consider the following:
$$\begin{align*}
  x_1^{y_1} &\geq x_2^{y_2} \\
  &\Leftrightarrow \\
  \log_{x_1} x_1^{y_1} &\geq \log_{x_1} x_2^{y_2} \\
  &\Leftrightarrow \\
  y_1 &\geq y_2 \log_{x_1} x_2
\end{align*}
$$
Without doing much calculation, we may be able to bound $\log_{x_1} x_2$ by two consecutive integers (recall, that $\log_{x_1} x_2$ is the exponent that when applied to $x_1$ results in the value $x_2$).  So, if we know that $n \leq \log_{x_1}x_2 \leq n+1$, it's easy to multiply $y_2$ by $n$ or $n+1$ and compare to $y_1$.  Unfortunately, if $y_1$ is very close to $ny_2$, we may not have a fine enough estimate to make the decision about the inequality.  In that event, you may have to use a calculator anyway.
A: Take logarithms.
If $\log(x_1^{y_1})>\log(x_1^{y_1})$ i.e. $y_1\log(x_1)>y_2\log(x_2)$, then $x_1^{y_1}>x_2^{y_2}$ and vice versa.
