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Question Number 152226 by mathdanisur last updated on 26/Aug/21

$${16}^{{\mathbf{x}}^2+\mathbf{y}}+{16}^{{\mathbf{y}}^2+\mathbf{x}}=1\Rightarrow \mathrm{x};\mathrm{y}=?$$

Commented by john_santu last updated on 26/Aug/21

$$\mathrm{x}=\mathrm{y}=-\frac{1}{2}$$

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