Hi, please help me on this question, I've been thinking 
about it for a long time. 

If loga(xy) = 3 and loga(x2y3) = 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've tried using the first log rule to do both equations
and came up with

loga(x) + loga(y) = 3 and 2·loga(x) + 3·loga(y) = 4
resectively but I don'tknow where to go from there.

That doesn't help

I also transformed it into an exponential equation

a3 = xy and a4 = x2y3 respectively and still have no clue
of what to do. 

This is a correct start.

You have to solve this system of equations:

a3 = xy
a4 = x2y3 

Solve the first one for x

x = a3/y

Substitute in the second one

a4 = x2y3 
a4 = (a3/y)2·y3

a4 = (a6/y2)·y3

a4 = a6y

a4 - a6y = 0

a4(1 - a2y) = 0

Using the zero-factor principle:

a4 = 0
or
1 - a2y = 0

We cannot have a4 = 0 because that would
make a = 0 and 0 cannot be the base of
a logarithm

So 1 - a2y = 0 
      -a2y = -1
         y = 1/a2

Substituting that in

x = a3/y

x = a3/(1/a2)

Invert and multiply

x = a3(a2/1)

x = a5     

So x - a5 = 0, or

x = a5

So the answer to the (a) part is

x = a5 and y = 1/a2 

To find the answer to (b), we eliminate a
from 

x = a5 and y = 1/a2

Clearing of fractions:

x = a5 and a2y = 1

We can make both equations have
a term in a10 by squaring both sides
of the first equation and raising both 
sides of the second to the 5th power:

x2 = a10 and  a10y5 = 1

Substitute x2 for a10 in the second

x2y5 = 1 

Edwin
AnlytcPhil@aol.com