Solve the following system of equations: $\begin{cases} 7^x - 16y = 0 \\ 4^x - 49y = 0 \end{cases}$
Given system of equations:
$7^x - 16y = 0$ $\;\;\; \cdots \; (1)$
$4^x - 49y = 0$ $\;\;\; \cdots \; (2)$
Multiplying equation $(1)$ with $49$ gives
$49 \times 7^x - 784 y = 0$
i.e. $\;$ $7^2 \times7^x - 784 y = 0$
i.e. $\;$ $7^{x + 2} = 784 y$ $\;\;\; \cdots \; (1a)$
Multiplying equation $(2)$ with $16$ gives
$16 \times 4^x - 784 y = 0$
i.e. $\;$ $4^2 \times4^x - 784 y = 0$
i.e. $\;$ $4^{x + 2} = 784 y$ $\;\;\; \cdots \; (2a)$
We have from equations $(1)$ and $(2)$
$7^{x+2} = 4^{x+2}$ $\;\;\; \cdots \; (3)$
i.e. $\;$ $\dfrac{7^{x+2}}{4^{x+2}} = 1$
i.e. $\;$ $\left(\dfrac{7}{4}\right)^{x+2} = 1$
i.e. $\;$ $x + 2 = \log_{\frac{7}{4}} 1$
i.e. $\;$ $x + 2 = 0$ $\implies$ $x = - 2$
Substituting the value of $x$ in equation $(1a)$ gives
$7^{-2 + 2} = 784 y$
i.e. $\;$ $7^0 = 784 y$
i.e. $\;$ $784 y = 1$ $\implies$ $y = \dfrac{1}{784}$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(-2, \; \dfrac{1}{784}\right) \right\}$