What is the answer to (2^{-2x})/(3^{2x)}=9^{-x} ?

The answer to (2^{-2x})/(3^{2x)}=9^{-x} is x=0

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(2^{-2x})/(3^{2x)}=9^{-x}

\frac{2 ^{- 2 x}}{3 ^{2 x}} = 9^{- x}

x = 0

Solution steps

\frac{2 ^{- 2 x}}{3 ^{2 x}} = 9^{- x}

Apply exponent rules

- 2 x ln (6) = - x ln (9)

Solve

- 2 x ln (6) = - x ln (9) : x = 0

x = 0

Verify Solutions:

x = 0

True

The solution is

x = 0

\frac{100}{2 + ( 0.75 ) ^{2 x}} = 45

286.93 (0.973 8^{x}) = 15

e^{x - 3} = x - 1

e^{4 - 4 x} = 9

e^{2 x} = 12 + 11 e^{x}

What is the answer to (2^{-2x})/(3^{2x)}=9^{-x} ?
The answer to (2^{-2x})/(3^{2x)}=9^{-x} is x=0
Solve for x: (2^{-2x})/(3^{2x)}=9^{-x}
The solution is x=0