# Elementary Ways to Solve System of Exponential Equation

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 :)

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 $$3^x$$

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

Comment:

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

$$(x, y)≈ (2.2, 1), (-2, 4.4)$$

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 $$(x, y) = (2, 3)$$.