<PRE><B>Question 39095</B>
&lt;pre&gt;&lt;font size = 4&gt;&lt;b&gt;Hi, please help me on this question, I&#39;ve been thinking 
about it for a long time. 

If log&lt;sub&gt;a&lt;/sub&gt;(xy) = 3 and log&lt;sub&gt;a&lt;/sub&gt;(x&lt;sup&gt;2&lt;/sup&gt;y&lt;sup&gt;3&lt;/sup&gt;) = 4, determine
 
a) the values of x and y in terms of a, using positive
indices.
b) an equation of x in terms of y

I&#39;ve tried using the first log rule to do both equations
and came up with

log&lt;sub&gt;a&lt;/sub&gt;(x) + log&lt;sub&gt;a&lt;/sub&gt;(y) = 3 and 2損og&lt;sub&gt;a&lt;/sub&gt;(x) + 3損og&lt;sub&gt;a&lt;/sub&gt;(y) = 4
resectively but I don&#39;tknow where to go from there.

&lt;i&gt;&lt;font color = &quot;indigo&quot;&gt;That doesn&#39;t help&lt;/i&gt;&lt;/font&gt;

I also transformed it into an exponential equation

a&lt;sup&gt;3&lt;/sup&gt; = xy and a&lt;sup&gt;4&lt;/sup&gt; = x&lt;sup&gt;2&lt;/sup&gt;y&lt;sup&gt;3&lt;/sup&gt; respectively and still have no clue
of what to do. 

&lt;i&gt;&lt;font color = &quot;indigo&quot;&gt;This is a correct start.

You have to solve this system of equations:

a&lt;sup&gt;3&lt;/sup&gt; = xy
a&lt;sup&gt;4&lt;/sup&gt; = x&lt;sup&gt;2&lt;/sup&gt;y&lt;sup&gt;3&lt;/sup&gt; 

Solve the first one for x

x = a&lt;sup&gt;3&lt;/sup&gt;/y

Substitute in the second one

a&lt;sup&gt;4&lt;/sup&gt; = x&lt;sup&gt;2&lt;/sup&gt;y&lt;sup&gt;3&lt;/sup&gt; 
a&lt;sup&gt;4&lt;/sup&gt; = (a&lt;sup&gt;3&lt;/sup&gt;/y)&lt;sup&gt;2&lt;/sup&gt;暄&lt;sup&gt;3&lt;/sup&gt;

a&lt;sup&gt;4&lt;/sup&gt; = (a&lt;sup&gt;6&lt;/sup&gt;/y&lt;sup&gt;2&lt;/sup&gt;)暄&lt;sup&gt;3&lt;/sup&gt;

a&lt;sup&gt;4&lt;/sup&gt; = a&lt;sup&gt;6&lt;/sup&gt;y

a&lt;sup&gt;4&lt;/sup&gt; - a&lt;sup&gt;6&lt;/sup&gt;y = 0

a&lt;sup&gt;4&lt;/sup&gt;(1 - a&lt;sup&gt;2&lt;/sup&gt;y) = 0

Using the zero-factor principle:

a&lt;sup&gt;4&lt;/sup&gt; = 0
or
1 - a&lt;sup&gt;2&lt;/sup&gt;y = 0

We cannot have a&lt;sup&gt;4&lt;/sup&gt; = 0 because that would
make a = 0 and 0 cannot be the base of
a logarithm

So 1 - a&lt;sup&gt;2&lt;/sup&gt;y = 0 
      -a&lt;sup&gt;2&lt;/sup&gt;y = -1
         y = 1/a&lt;sup&gt;2&lt;/sup&gt;

Substituting that in

x = a&lt;sup&gt;3&lt;/sup&gt;/y

x = a&lt;sup&gt;3&lt;/sup&gt;/(1/a&lt;sup&gt;2&lt;/sup&gt;)

Invert and multiply

x = a&lt;sup&gt;3&lt;/sup&gt;(a&lt;sup&gt;2&lt;/sup&gt;/1)

x = a&lt;sup&gt;5&lt;/sup&gt;     

So x - a&lt;sup&gt;5&lt;/sup&gt; = 0, or

x = a&lt;sup&gt;5&lt;/sup&gt;

So the answer to the (a) part is

x = a&lt;sup&gt;5&lt;/sup&gt; and y = 1/a&lt;sup&gt;2&lt;/sup&gt; 

To find the answer to (b), we eliminate a
from 

x = a&lt;sup&gt;5&lt;/sup&gt; and y = 1/a&lt;sup&gt;2&lt;/sup&gt;

Clearing of fractions:

x = a&lt;sup&gt;5&lt;/sup&gt; and a&lt;sup&gt;2&lt;/sup&gt;y = 1

We can make both equations have
a term in a&lt;sup&gt;10&lt;/sup&gt; by squaring both sides
of the first equation and raising both 
sides of the second to the 5th power:

x&lt;sup&gt;2&lt;/sup&gt; = a&lt;sup&gt;10&lt;/sup&gt; and  a&lt;sup&gt;10&lt;/sup&gt;y&lt;sup&gt;5&lt;/sup&gt; = 1

Substitute x&lt;sup&gt;2&lt;/sup&gt; for a&lt;sup&gt;10&lt;/sup&gt; in the second

x&lt;sup&gt;2&lt;/sup&gt;y&lt;sup&gt;5&lt;/sup&gt; = 1 

Edwin
AnlytcPhil@aol.com&lt;/pre&gt;&lt;/font&gt;&lt;i&gt;&lt;/font&gt;
	
	

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