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Is there any elementary way (or using Lambert-W maybe) to solve this system of the exponential equation: $$ \begin{cases} 3^{x+y}+2^{y-1}=23, \\ 3^{2x-1}+2^{y+1}=43. \end{cases} $$

I have tried to eliminate the exponent of 2 but it gets me $$ 12 \cdot 3^{x + y} + 3^{2x} = 405 $$ which is more complicated.

I have also tried to substitute $ 3^x = u $ and $ 2^y = v $ but there is still $ 3^y $.

Any advice is welcome (it's okay to use non-elementary method). Thanks :)

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Hint:

$$(3^x)^2+(12\cdot3^y)3^x-405=0$$

The discriminant is $$(12\cdot3^y)^2+4\cdot405=16\cdot3^{2y+2}+3^4\cdot20=4\cdot3^2(4\cdot9^y+45)$$

For rational $3^x,$ we need $$(2\cdot3^y)^2+45$$ to be perfect square $=d^2, d\ge0$(say)

$$\implies45=d^2-(2\cdot3^y)^2=(d+2\cdot3^y)(d-2\cdot3^y)\le(d+2\cdot3^y)^2$$

$$\implies d+2\cdot3^y\ge\sqrt{45}>6$$

Again, $d+2\cdot3^y$ must divide $45,$ hence can be one of $$\{9,15,45\}$$

From here we can find $3^y$ and $d$ and hence $3^x$

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    $\begingroup$ Why must $3^x$ be rational? $\endgroup$ – player3236 Nov 9 '20 at 8:36
  • $\begingroup$ (x, y)=(-2, 4.4), (2.2, 1) $\endgroup$ – sirous Nov 9 '20 at 8:45
  • $\begingroup$ @player3236, I said if $3^x$ is rational $\endgroup$ – lab bhattacharjee Nov 9 '20 at 9:18
  • $\begingroup$ Just making sure that it could be irrational. Secondly, why must $d+2\cdot 3^y$ be an integer? $\endgroup$ – player3236 Nov 9 '20 at 9:29
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Comment:

Finding by plotting the equations, using Wolfram we get following figure:

enter image description here

$(x, y)≈ (2.2, 1), (-2, 4.4)$

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It turns out that my Professor is making a mistake. The correct system of equations is $$ \begin{cases} 3^{x + y} + 2^{y - 1} = 239, \\ 3^{2x - 1} + 2^{y + 1} = 43. \end{cases} $$

Answer: We know that $ 239 = 243 - 4 = 3^5 - 2^2 $ and $ 43 = 27 + 16 = 3^3 + 2^4 $, so we have new system of equations that satisfy $$ \begin{cases} x + y = 5, \\ y - 1 = 2, \\ 2x - 1 = 3, \\ y + 1 = 4. \end{cases} $$ Which only satisfied for $ (x, y) = (2, 3) $.

And also, thanks for your answer. I really appreciate it.

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    $\begingroup$ This is what I found from second equation but it did not satisfy the first equation.Plotting gives :(x,y)=(2,3), (1.5, 4.5). so you found only one solution. $\endgroup$ – sirous Nov 9 '20 at 10:24

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