# Exponential equation

1. Apr 27, 2008

### Physicsissuef

1. The problem statement, all variables and given/known data
Solve this system of equations:

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

2. Relevant equations

3. The attempt at a solution

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

2. Apr 27, 2008

### HallsofIvy

Staff Emeritus
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.

3. Apr 27, 2008

### Physicsissuef

This system of equations have no solution?

4. Apr 27, 2008

### HallsofIvy

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

5. Apr 27, 2008

### Physicsissuef

The actual problem was:

Here.

But I simplify it to the one above.

6. Apr 27, 2008

### 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.

7. Apr 27, 2008

### Physicsissuef

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

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

8. Apr 27, 2008

### tmclary

Use Hall of Ivy substitution

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

9. Apr 27, 2008

### Physicsissuef

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

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

But where I will go out of herE?