User:Kamek77

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[edit] New Math

cuberoot 2^2-x = 2^x^2

(\sqrt[3]{2})^{2-x} = 2^{x^2}
(2^{\frac{1}{3}})^{2-x} = 2^{x^2}
2^{\frac{1}{3}\left({2-x}\right)} = 2^{x^2}

Since the bases are equal, the exponents are equal.

Thus \frac{1}{3}\left({2-x}\right) = {x^2}
This can be solved using either the quadratic formula or by completing the square.

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