Solve the following system of equations: $\begin{cases} 3^{2x} - 2^y = 77 \\ 3^x - 2^{\frac{y}{2}} = 7 \end{cases}$
Given system of equations:
$3^{2x} - 2^y = 77$ $\;\;\; \cdots \; (1)$
$3^x - 2^{\frac{y}{2}} = 7$ $\;\;\; \cdots \; (2)$
Equation $(1)$ can be written as
$\left(3^x\right)^2 - \left(2^{\frac{y}{2}}\right)^2 = 77$
i.e. $\;$ $\left(3^x + 2 ^{\frac{y}{2}}\right) \left(3^x - 2^{\frac{y}{2}}\right) = 77$
i.e. $\;$ $\left(3^x + 2^{\frac{y}{2}}\right) \times 7 = 77$ $\;\;\;$ [in view of equation $(2)$]
i.e. $\;$ $3^x + 2^{\frac{y}{2}} = 11$ $\;\;\; \cdots \; (3)$
Adding equations $(2)$ and $(3)$ gives
$2 \times 3^x = 18$
i.e. $\;$ $3^x = 9 = 3^2$
$\implies$ $x = 2$
Substituting the value of $x$ in equation $(3)$ gives
$3^2 + 2^{\frac{y}{2}} = 11$
i.e. $\;$ $2^{\frac{y}{2}} = 2 = 2^1$
$\implies$ $\dfrac{y}{2} = 1$ $\implies$ $y = 2$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(2, 2\right) \right\}$