What is the answer to 2^{x^2+x}=4^{x+1} ?

The answer to 2^{x^2+x}=4^{x+1} is x=2,x=-1

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2^{x^2+x}=4^{x+1}

2^{x^{2} + x} = 4^{x + 1}

x = 2, x = - 1

Solution steps

2^{x^{2} + x} = 4^{x + 1}

Apply exponent rules

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

Solve

x^{2} + x = 2 (x + 1) : x = 2, x = - 1

x = 2, x = - 1

1 6^{2 x + 1} = 3^{3 x - 4}

(- 4)^{x} = \frac{1}{16}

1 = 3^{x}

e^{2 x} + e^{x} - 6 = 0

2^{x} + 2^{3 - x} - 9 = 0

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