SOLUTION: How can i solve this system of equations? {{{system(matrix(2,1,"", 3^y*64^(1/x)=36),matrix(2,1,"", 5^y*512^(1/x)=200)) }}}

Algebra ->  Equations -> SOLUTION: How can i solve this system of equations? {{{system(matrix(2,1,"", 3^y*64^(1/x)=36),matrix(2,1,"", 5^y*512^(1/x)=200)) }}}      Log On


   

Question 903215: How can i solve this system of equations?
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!


The 1%2Fx is cumbersome, so let's substitute  1%2Fx=u

system%283%5Ey%2A64%5Eu=36%2C5%5Ey%2A512%5Eu=200%29

Write 64 as 2%5E6,
Write 36 as 2%5E2%2A3%5E2
Write 512 as 2%5E9
Write 200 as 2%5E3%2A5%5E2

system%283%5Ey%2A%282%5E6%29%5Eu=2%5E2%2A3%5E2%2C5%5Ey%2A%282%5E9%29%5Eu=2%5E3%2A5%5E2%29

system%283%5Ey%2A2%5E%286u%29=2%5E2%2A3%5E2%2C5%5Ey%2A2%5E%289u%29=2%5E3%2A5%5E2%29

Divide the first equation through by 3%5E2%2A2%5E%286u%29
Divide the second equation through hy 5%5E2%2A2%5E%289u%29

, 

Subtracting exponents of like bases:

system%283%5E%28y-2%29=2%5E%282-6u%29%2C5%5E%28y-2%29=2%5E%283-9u%29%29

Take out common factors in exponents on the right:

system%283%5E%28y-2%29=2%5E%282%281-3u%29%29%2C5%5E%28y-2%29=2%5E%283%281-3u%29%29%29

Write powers of 2 as 2nd and 3rd powers of 2%5E%281-3u%29

system%283%5E%28y-2%29=%282%5E%281-3u%29%29%5E2%2C5%5E%28y-2%29=%282%5E%281-3u%29%29%5E3%29

To make the right sides equal raise both sides of the first
equation to the 3rd power and raise both sides of the second
equation to the second power:








Divide equals by equals.  Since the right sides are equal, the
right side will be just 1.

%283%5E%283%28y-2%29%29%29%2F%285%5E%282%28y-2%29%29%29=1

The left side can be written as

%283%5E3%2F5%5E2%29%5E%28y-2%29=1

The only exponent of a base (other than 1) that equals 1 is 0,
therefore

y-2=0

y=2

Substituting in

3%5Ey%2A2%5E%286u%29=2%5E2%2A3%5E2

3%5E2%2A2%5E%286u%29=2%5E2%2A3%5E2

2%5E%286u%29=2%5E2

Equating the exponents of 2

6u+=+2

u=2%2F6=1%2F3

1%2Fx=u=1%2F3

x=3

Solution: (x,y) = (3,2) 

Edwin