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

16^(x^2 +y)  + 16^(y^2 +x)  = 1  ⇒  x;y=?

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

Commented by john_santu last updated on 26/Aug/21

x=y=−(1/2)

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

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