What is the answer to solvefor x,2^{x+1}= 1/(2^{2x+3)} ?

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

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solvefor x,2^{x+1}= 1/(2^{2x+3)}

solve for

x, 2^{x + 1} = \frac{1}{2 ^{2 x + 3}}

x = - \frac{4}{3}

Decimal Notation

x = - 1.33333 \dots

Solution steps

2^{x + 1} = \frac{1}{2 ^{2 x + 3}}

Apply exponent rules

x + 1 = - 2 x - 3

Solve

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

x = - \frac{4}{3}

Verify Solutions:

x = - \frac{4}{3}

True

The solution is

x = - \frac{4}{3}

3 2^{x} = (\frac{1}{128})^{x - 4}

4^{x + 1} + 4^{x + 2} - 320 = 0

a x \cdot e^{b x}

e^{4 x - 4} = 9

1 0^{2 x} = 2

What is the answer to solvefor x,2^{x+1}= 1/(2^{2x+3)} ?
The answer to solvefor x,2^{x+1}= 1/(2^{2x+3)} is x=-4/3