Solve the system $3^x-2^{y^2}=77$, $3^{\frac{x}{3}}-2^{\frac{y^2}{2}}=7$ in $\mathbb{R}$

Question

I had to solve the similar system $3^x-2^{y^2}=77,\; 3^{\frac{x}{2}}-2^{\frac{y^2}{2}}=7$ before, which can be solved like this: $$3^{\frac{x}{2}}-2^{\frac{y^2}{2}}=7 \implies 2^{\frac{y^2}{2}}=3^{\frac{x}{2}}-7$$ $$2^{\frac{y^2}{2}}=3^{\frac{x}{2}}-7 \implies 2^{y^2}=(3^{\frac{x}{2}}-7)^2$$ $$2^{y^2}=(3^{\frac{x}{2}}-7)^2 \implies 2^{y^2}=3^{x}-14\cdot3^{\frac{x}{2}}+49$$

So putting that in the first equation: $$3^x-2^{y^2}=77 \implies 3^x-3^{x}+14\cdot3^{\frac{x}{2}}-49=77$$ $$\mathbf{3^x-3^{x}+14\cdot3^{\frac{x}{2}}-49=77 \implies 14\cdot3^{\frac{x}{2}}=126}$$ $$14\cdot3^{\frac{x}{2}}=126 \implies 3^{\frac{x}{2}}=9$$ $$3^{\frac{x}{2}}=9 \implies x = 4$$

And putting the value of $x$ in the second equation we get $y$: $$3^{\frac{x}{2}}-2^{\frac{y^2}{2}}=7 \implies 3^{2}-2^{\frac{y^2}{2}}=7$$ $$3^{2}-2^{\frac{y^2}{2}}=7 \implies 2^{\frac{y^2}{2}}=2$$ $$2^{\frac{y^2}{2}}=2 \implies y^2 = 2$$ $$y^2 = 2 \implies y = \sqrt{2}$$

So I tried solving the system in the question title with the same strategy until the step in bold, which gave me $3^x - 3^{\frac{2x}{3}}+14\cdot3^{\frac{x}{3}}=126$, and I have no idea how to progress further.

Answer 1

If $u=3^{\frac x 2}$ and $v=2^{\frac {y^2}2}$ then $$\left \{ \begin{split} u^3-v^2&=77\\ u-v&=7 \end{split} \right.$$ So $v=u-7$ and thus $u^3-u^2+14u-126=0$.

Solving the latter gives you 3 roots (1 real and 2 complex), and you can find $u$ and $v$, and get back to $x$ and $y$.

Answer 2

Using the hyperbolic method when there is only one real root as for$$u^3-u^2+14u-126=0$$ the solution is not ugly at all. It write $$u=\frac{1}{3} \left(1+2 \sqrt{41} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{1639}{41 \sqrt{41}}\right)\right)\right)$$ while using Cardano method, it write $$\frac{1}{3} \left(1-\frac{41}{\sqrt[3]{1639+3 \sqrt{306138}}}+\sqrt[3]{1639+3 \sqrt{306138}}\right)$$

Which one do you prefer ?