What is the answer to 2^{2x}+8(2^x)+7=0 ?

The answer to 2^{2x}+8(2^x)+7=0 is No Solution for x\in\mathbb{R}

Solve for x: 2^{2x}+8(2^x)+7=0

The solution is No Solution for x\in\mathbb{R}

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2^{2 x} + 8 (2^{x}) + 7 = 0

No Solution for x \in R

Solution steps

2^{2 x} + 8 (2^{x}) + 7 = 0

Apply exponent rules

(2^{x})^{2} + 8 \cdot 2^{x} + 7 = 0

Rewrite the equation with

2^{x} = u

(u)^{2} + 8 u + 7 = 0

Solve

u^{2} + 8 u + 7 = 0 : u = - 1, u = - 7

u = - 1, u = - 7

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = - 1 :

No Solution for

x \in R

Solve

2^{x} = - 7 :

No Solution for

x \in R

No Solution for x \in R

4^{x} = 23

e^{5 x} = 6

(\frac{1}{2})^{3 x} = 8

x, 2 t = e^{x} - e^{- x}

e^{2} x = 2 e^{x}

What is the answer to 2^{2x}+8(2^x)+7=0 ?
The answer to 2^{2x}+8(2^x)+7=0 is No Solution for x\in\mathbb{R}
Solve for x: 2^{2x}+8(2^x)+7=0
The solution is No Solution for x\in\mathbb{R}