Solve the following system of equations: $\begin{cases} 4^{x+y} = 128 \\ 5^{3x-2y-3} = 1 \end{cases}$
Given system of equations:
$4^{x+y} = 128$ $\;\;\; \cdots \; (1)$
$5^{3x-2y-3} = 1$ $\;\;\; \cdots \; (2)$
Equation $(1)$ $\implies$ $x + y = \log_4 128$
i.e. $\;$ $x + y = \log_4 2^7$
i.e. $\;$ $x + y = 7 \log_4 2$
i.e. $\;$ $x + y = 7 \log_4 4^{\frac{1}{2}}$
i.e. $\;$ $x + y = \dfrac{7}{2} \log_4 4$
i.e. $\;$ $x + y = \dfrac{7}{2}$
i.e. $\;$ $2x + 2y = 7$ $\;\;\; \cdots \; (3)$
Equation $(2)$ $\implies$ $3x - 2y - 3 = \log_5 1$
i.e. $\;$ $3x - 2y - 3 = 0$
i.e. $\;$ $3x - 2y = 3$ $\;\;\; \cdots \; (4)$
Solving equations $(3)$ and $(4)$ simultaneously [adding equations $(3)$ and $(4)$] gives
$5x = 10$ $\implies$ $x = 2$
Substituting the value of $x$ in equation $(3)$ gives
$y = \dfrac{7 - 4}{2} = \dfrac{3}{2}$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(2, \dfrac{3}{2}\right) \right\}$