Solving 3^(xy)=2^(yx) and 12^(xx)=3^(y4)

[Physicsissuef]

Homework Statement

Solve this system of equations:

$$3^xy=2^yx$$
$$12^xx=3^y4$$

Homework Equations

The Attempt at a Solution

I was solving and came up, till here:
$$x=\frac{3^xy}{2^y}$$
$$6^y4=36^x$$

Please help. Thanks.

 

[HallsofIvy]

Actually I'm surprised you were able to get that far! Most equations that involve variables both "inside" and "outside" transcendental functions cannot be solved in terms of elementary functions.

Once you are at
$$6^y4= 36^x= (6^2)^x= 6^{2x}$$
You can take the logarithm of both sides:
$$y ln(6)+ ln 4= 2x ln 5$$
Where you would go from there, I have no idea.

 

[Physicsissuef]

This system of equations have no solution?

 

[HallsofIvy]

I didn't say that. It said it might not be possible to solve it using elementary functions.

 

[Physicsissuef]

The actual problem was:

"[URL 0012.jpg"]Here.[/URL]

But I simplify it to the one above.

 

[symbolipoint]

Try using change of base of the logarithm functions first, and then try. Put then into either base 2 or base 3. I have not tried this in your exercise but believe it's worth trying.

 

[Physicsissuef]

I tried on several ways and it didn't worked.

btw- on the first post should be:
$$6^y4=36^xy$$

 

[tmclary]

Use Hall of Ivy substitution

in the above equation: (6^2x)y=(6^y)4.

 

[Physicsissuef]

Maybe
$$log_66^y4=log_66^2xy$$

$$y+log_64=2x+log_6y$$

But where I will go out of herE?