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

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

Solve for x: 2^{2x+2}=3-2^x

The solution is x=(ln(3/4))/(ln(2))

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

x = \frac{ln ( \frac{3}{4} )}{ln ( 2 )}

Solution steps

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

Apply exponent rules

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

Rewrite the equation with

2^{x} = u

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

Solve

u^{2} \cdot 2^{2} = 3 - u : u = \frac{3}{4}, u = - 1

u = \frac{3}{4}, u = - 1

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = \frac{3}{4} : x = \frac{ln ( \frac{3}{4} )}{ln ( 2 )}

Solve

2^{x} = - 1 :

No Solution for

x \in R

x = \frac{ln ( \frac{3}{4} )}{ln ( 2 )}

82 = 2^{x}

4^{x - 2} = 13

4 = e^{2 x}

9^{5 x + 5} = 8^{9 x - 5}

e^{1 - x} = e^{1} e^{- x}

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