What is the answer to 2^{2x+2}=9(2^x)^{-1} ?

The answer to 2^{2x+2}=9(2^x)^{-1} is x=(2ln(3)-2ln(2))/(3ln(2))

Solve for x: 2^{2x+2}=9(2^x)^{-1}

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

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2^{2 x + 2} = 9 (2^{x})^{- 1}

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

Decimal Notation

x = 0.38997 \dots

Solution steps

2^{2 x + 2} = 9 (2^{x})^{- 1}

Apply exponent rules

(2 x + 2) ln (2) = ln (9) + x (- 1) ln (2)

Solve

(2 x + 2) ln (2) = ln (9) + x (- 1) ln (2) : x = \frac{2 ln ( 3 ) - 2 ln ( 2 )}{3 ln ( 2 )}

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

4^{x} = \frac{1}{2}

y, x^{y} - y^{x} = 0

64 = 2^{x - 1}

3 \cdot 2^{x} = 1

3^{x + 4} = 243

What is the answer to 2^{2x+2}=9(2^x)^{-1} ?
The answer to 2^{2x+2}=9(2^x)^{-1} is x=(2ln(3)-2ln(2))/(3ln(2))
Solve for x: 2^{2x+2}=9(2^x)^{-1}
The solution is x=(2ln(3)-2ln(2))/(3ln(2))