What is the answer to 4^{4x^2}= 1/(16^{2x-3)} ?

The answer to 4^{4x^2}= 1/(16^{2x-3)} is x=(-1+sqrt(7))/2 ,x=-(1+sqrt(7))/2

Solve for x: 4^{4x^2}= 1/(16^{2x-3)}

The solution is x=(-1+sqrt(7))/2 ,x=-(1+sqrt(7))/2

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4^{4 x^{2}} = \frac{1}{1 6 ^{2 x - 3}}

x = \frac{- 1 + 7}{2}, x = - \frac{1 + 7}{2}

Decimal Notation

x = 0.82287 \dots, x = - 1.82287 \dots

Solution steps

4^{4 x^{2}} = \frac{1}{1 6 ^{2 x - 3}}

Apply exponent rules

4 x^{2} = 2 (- 2 x + 3)

Solve

4 x^{2} = 2 (- 2 x + 3) : x = \frac{- 1 + 7}{2}, x = - \frac{1 + 7}{2}

x = \frac{- 1 + 7}{2}, x = - \frac{1 + 7}{2}

Verify Solutions:

x = \frac{- 1 + 7}{2}

True

, x = - \frac{1 + 7}{2}

True

The solutions are

x = \frac{- 1 + 7}{2}, x = - \frac{1 + 7}{2}

\frac{4}{21} = e^{- 203 t}

2^{x} + 1 - \frac{3}{4} 2^{- x} = 0

9 \cdot 1 0^{x} - 81 = 0

(x + 2)^{3 x - 2} = (x + 2)^{x - 4}

x = 2 e^{- x}

What is the answer to 4^{4x^2}= 1/(16^{2x-3)} ?
The answer to 4^{4x^2}= 1/(16^{2x-3)} is x=(-1+sqrt(7))/2 ,x=-(1+sqrt(7))/2
Solve for x: 4^{4x^2}= 1/(16^{2x-3)}
The solution is x=(-1+sqrt(7))/2 ,x=-(1+sqrt(7))/2