What is the answer to (\sqrt[3]{2})^{2-x}=2^{x^2} ?

The answer to (\sqrt[3]{2})^{2-x}=2^{x^2} is x= 2/3 ,x=-1

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(\sqrt[3]{2})^{2-x}=2^{x^2}

x = \frac{2}{3}, x = - 1

Decimal Notation

x = 0.66666 \dots, x = - 1

Solution steps

Apply exponent rules

\frac{1}{3} (2 - x) = x^{2}

Solve

\frac{1}{3} (2 - x) = x^{2} : x = \frac{2}{3}, x = - 1

x = \frac{2}{3}, x = - 1

2^{3 - 3 x} = 64

- 2^{x + 1} + 5 \cdot 2^{x} = 28

1 0^{a} = 1000000

2^{2 x + 2} = 3 - 2^{x}

82 = 2^{x}

What is the answer to (\sqrt[3]{2})^{2-x}=2^{x^2} ?
The answer to (\sqrt[3]{2})^{2-x}=2^{x^2} is x= 2/3 ,x=-1
Solve for x: (\sqrt[3]{2})^{2-x}=2^{x^2}
The solution is x= 2/3 ,x=-1