How do you solve #8^(xy)=3^x# and #8^y=2^(x3)#?
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To solve the system of equations:

(8^{(xy)} = 3^x)

(8^y = 2^{(x3)})
First, recognize that (8 = 2^3) and rewrite the equations in terms of the base 2:

((2^3)^{(xy)} = 3^x) becomes (2^{3(xy)} = 3^x)

((2^3)^y = 2^{(x3)}) becomes (2^{3y} = 2^{(x3)})
From equation 2) (2^{3y} = 2^{(x3)}), since the bases are the same, the exponents must be equal:
3y = x  3
Rearrange to solve for x:
x = 3y + 3
Now, substitute x in the first equation:
(2^{3(xy)} = 3^x)
Substitute x = 3y + 3:
(2^{3((3y+3)y)} = 3^{(3y+3)})
Simplify the exponent on the left side:
(2^{3(2y+3)} = 3^{(3y+3)})
Notice that directly solving for y or x from here involves dealing with logarithms that do not easily simplify due to the mismatch of bases 2 and 3. However, we realize that solving for x and y in terms of simple algebraic manipulation directly from this point is complex due to the base mismatch. The key is recognizing the relationship and expressing one variable in terms of the other, which has been done (x in terms of y). For specific numerical solutions, one would typically resort to numerical methods or graphing techniques given the transcendental nature of the equation (involving exponential functions with different bases).
For algebraic manipulation, the approach shown sets up the relationship between x and y. The equations indicate how these variables are intertwined through their exponents, but without specific values to work towards or additional constraints, we're at the limit of simple algebraic manipulation. Typically, further progress would require numerical solutions or approximation methods, especially given the base discrepancy (2 and 3) that doesn't lend itself to a neat, analytical solution.
For the sake of clarity and completion, while the analytical steps taken demonstrate the approach to manipulate exponential equations into a form where one variable is expressed in terms of the other, practical resolution of this specific set to get exact values for x and y under normal algebraic means isn't feasible. One might use a graphing approach or numerical algorithms (like NewtonRaphson for systems of equations) to find approximate solutions for x and y.
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