What is the answer to 2^{-2(4+x)}=sqrt(2^{x-1)} ?

The answer to 2^{-2(4+x)}=sqrt(2^{x-1)} is x=-3

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2^{-2(4+x)}=sqrt(2^{x-1)}

2^{- 2 (4 + x)} = 2^{x - 1}

x = - 3

Solution steps

2^{- 2 (4 + x)} = 2^{x - 1}

Apply exponent rules

- 2 (4 + x) = (x - 1) \frac{1}{2}

Solve

- 2 (4 + x) = (x - 1) \frac{1}{2} : x = - 3

x = - 3

e^{5 - 4 x} = 2

e^{5 - 4 x} = 4

(\frac{2}{3})^{x} = 8

1 1^{x} = 2

2^{2 x - 1} + 8 = 5 (2^{x})

What is the answer to 2^{-2(4+x)}=sqrt(2^{x-1)} ?
The answer to 2^{-2(4+x)}=sqrt(2^{x-1)} is x=-3
Solve for x: 2^{-2(4+x)}=sqrt(2^{x-1)}
The solution is x=-3