SOLUTION: Hi, please help me on this question, I've been thinking about it for a long time. If {{{log(base a)(xy) = 3}}} and {{{log(base a)(x^2y^3) = 4}}}, determine a) the values of

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Hi, please help me on this question, I've been thinking about it for a long time. If {{{log(base a)(xy) = 3}}} and {{{log(base a)(x^2y^3) = 4}}}, determine a) the values of       Log On


   

Question 39095: Hi, please help me on this question, I've been thinking about it for a long time.
If log%28base+a%29%28xy%29+=+3 and log%28base+a%29%28x%5E2y%5E3%29+=+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 log%28base+a%29x+%2B+log%28base+a%29y+=+3 and 2log%28base+a%29x+%2B+3log%28base+a%29y+=+4 resectively but I don't know where to go from there. I also transformed it into an exponential equation a%5E3+=xy and a%5E4+=+x%5E2y%5E3 respectively and still have no clue of what to do.
Thank you.
Found 2 solutions by fractalier, AnlytcPhil:
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
Yes, do it the second way, where you had
a^3 = xy and
a^4 = (x^2)(y^3)
Now solve the first for x and plug it in to the second
x = (a^3)/y and then
a^4 = [(a^6)/(y^2)](y^3) so that
y = a^(-2)
and since
a^3 = xy we have
x = a^5
Now from y = a^(-2) we have
a^2 = 1/y = y^(-1) and
x = a^5 = y^(-5/2)
and we're done.

Answer by AnlytcPhil(1810) About Me  (Show Source):
You can put this solution on YOUR website!
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