What is the answer to 64^{2^{1-x}}=8^{2^{x+1}} ?

The answer to 64^{2^{1-x}}=8^{2^{x+1}} is x= 1/2

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64^{2^{1-x}}=8^{2^{x+1}}

6 4^{2^{1 - x}} = 8^{2^{x + 1}}

x = \frac{1}{2}

Decimal Notation

x = 0.5

Solution steps

6 4^{2^{1 - x}} = 8^{2^{x + 1}}

Apply exponent rules

- x + 2 = x + 1

Solve

- x + 2 = x + 1 : x = \frac{1}{2}

x = \frac{1}{2}

x^{2} e^{x} - x e^{x} - 2 e^{x} = 0

2^{∣ A ∣} = 2^{4}

2^{n - 3} = 8^{n}

2 7^{2} = 3^{x}

(2 x^{2} + 4 x - 16) \cdot 3^{x} = 0

What is the answer to 64^{2^{1-x}}=8^{2^{x+1}} ?
The answer to 64^{2^{1-x}}=8^{2^{x+1}} is x= 1/2
Solve for x: 64^{2^{1-x}}=8^{2^{x+1}}
The solution is x= 1/2