<PRE><B>Question 65366</B>
&lt;pre&gt;&lt;font size = 5&gt;&lt;b&gt;
&lt;u&gt; 16x&lt;sup&gt;5&lt;/sup&gt;y&lt;sup&gt;4&lt;/sup&gt; &lt;/u&gt;
 32xy&lt;sup&gt;7&lt;/sup&gt;

Give the exponent 1 to the letter x in the bottom

&lt;u&gt; 16x&lt;sup&gt;5&lt;/sup&gt;y&lt;sup&gt;4&lt;/sup&gt; &lt;/u&gt;
 32x&lt;sup&gt;1&lt;/sup&gt;y&lt;sup&gt;7&lt;/sup&gt;

Cancel the 16 and the 32, putting 2
below the 32 since 32÷16 = 2, and
putting 1 above the 16, since 16÷16 = 1

     1
&lt;u&gt; &lt;s&gt;16&lt;/s&gt;x&lt;sup&gt;5&lt;/sup&gt;y&lt;sup&gt;4&lt;/sup&gt; &lt;/u&gt;
 &lt;s&gt;32&lt;/s&gt;x&lt;sup&gt;1&lt;/sup&gt;y&lt;sup&gt;7&lt;/sup&gt;
     2

So now you have

&lt;u&gt; 1x&lt;sup&gt;5&lt;/sup&gt;y&lt;sup&gt;4&lt;/sup&gt; &lt;/u&gt;
 2x&lt;sup&gt;1&lt;/sup&gt;y&lt;sup&gt;7&lt;/sup&gt;

Now use the rule for subtracting exponents:

When two exponential with the same base
appear as a factor of the numerator and
the denominator:

Subtract the smaller exponent from the larger
exponent and place 

(a) the resulting exponent with the common base
in the numerator if that&#39;s where the larger 
exponent was previously and eliminate it from 
the denominator.   Or,

(b) the resulting exponent with the common base
in the denominator if that&#39;s where the larger 
exponent was previously and eliminate it from 
the numerator. 

So here we have a case of each

&lt;u&gt; 1x&lt;sup&gt;5&lt;/sup&gt;y&lt;sup&gt;4&lt;/sup&gt; &lt;/u&gt;
 2x&lt;sup&gt;1&lt;/sup&gt;y&lt;sup&gt;7&lt;/sup&gt;

We subtract the exponent of x&lt;sup&gt;1&lt;/sup&gt; from the exponent 
of x&lt;sup&gt;5&lt;/sup&gt;, and get x&lt;sup&gt;4&lt;/sup&gt;, and we place x&lt;sup&gt;4&lt;/sup&gt; in the numerator
because the larger exponent 5 was in the numerator,
and we eliminate x from the denominator:

&lt;u&gt; 1x&lt;sup&gt;4&lt;/sup&gt;y&lt;sup&gt;4&lt;/sup&gt; &lt;/u&gt;
 2y&lt;sup&gt;7&lt;/sup&gt;

We subtract the exponent of y&lt;sup&gt;4&lt;/sup&gt; from the exponent 
of y&lt;sup&gt;7&lt;/sup&gt;, and get y&lt;sup&gt;3&lt;/sup&gt;, and we place y&lt;sup&gt;3&lt;/sup&gt; in the denominator
because the larger exponent 7 was in the denominator,
and we eliminate y from the numerator:


&lt;u&gt; 1x&lt;sup&gt;4&lt;/sup&gt; &lt;/u&gt;
 2y&lt;sup&gt;3&lt;/sup&gt;

Now we can erase the 1 coefficient in the top

&lt;u&gt;  x&lt;sup&gt;4&lt;/sup&gt; &lt;/u&gt;
 2y&lt;sup&gt;3&lt;/sup&gt;

Edwin&lt;/pre&gt;</PRE>