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

The answer to 2^{2x+1}=3(2^x)-1 is x=0,x=-1

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2^{2x+1}=3(2^x)-1

2^{2 x + 1} = 3 (2^{x}) - 1

x = 0, x = - 1

Solution steps

2^{2 x + 1} = 3 (2^{x}) - 1

Apply exponent rules

(2^{x})^{2} \cdot 2^{1} = 3 \cdot 2^{x} - 1

Rewrite the equation with

2^{x} = u

(u)^{2} \cdot 2^{1} = 3 u - 1

Solve

u^{2} \cdot 2^{1} = 3 u - 1 : u = 1, u = \frac{1}{2}

u = 1, u = \frac{1}{2}

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 1 : x = 0

Solve

2^{x} = \frac{1}{2} : x = - 1

x = 0, x = - 1

2 7^{x} = 8 1^{x}

x = e^{\frac{1}{2} x - 2}

5^{x^{2} - 1} = 1

x, 35 - 12 e^{5 - 4 x} = 0

7^{w - 2} = 49

What is the answer to 2^{2x+1}=3(2^x)-1 ?
The answer to 2^{2x+1}=3(2^x)-1 is x=0,x=-1
Solve for x: 2^{2x+1}=3(2^x)-1
The solution is x=0,x=-1