4
$\begingroup$

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 :)

$\endgroup$
4
$\begingroup$

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 $3^x$

$\endgroup$
4
  • 1
    $\begingroup$ 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
1
$\begingroup$

Comment:

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

enter image description here

$(x, y)≈ (2.2, 1), (-2, 4.4)$

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.