Exponential Equation System: Solving 3^xy=2^yx and 12^xx=3^y4

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The forum discussion focuses on solving the exponential equation system defined by 3^xy = 2^yx and 12^xx = 3^y4. Participants highlight the complexity of these equations, noting that they may not be solvable using elementary functions. The discussion suggests using logarithmic transformations and change of base techniques to approach the problem, specifically recommending the use of base 2 or base 3 logarithms. The conversation emphasizes the need for advanced algebraic manipulation and substitution methods to explore potential solutions.

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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.
 
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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.
 
This system of equations have no solution?
 
I didn't say that. It said it might not be possible to solve it using elementary functions.
 
The actual problem was:

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

But I simplify it to the one above.
 
Last edited by a moderator:
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.
 
I tried on several ways and it didn't worked.

btw- on the first post should be:
<br /> 6^y4=36^xy<br />
 
Use Hall of Ivy substitution

in the above equation: (6^2x)y=(6^y)4.
 
Maybe
log_66^y4=log_66^2xy

y+log_64=2x+log_6y

But where I will go out of herE?
 

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