What is the answer to (3^2x)/(3^{-7)}=3^{-3} ?

The answer to (3^2x)/(3^{-7)}=3^{-3} is x= 1/(3^{12)}

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(3^2x)/(3^{-7)}=3^{-3}

\frac{3 ^{2} x}{3 ^{- 7}} = 3^{- 3}

x = \frac{1}{3 ^{12}}

Decimal Notation

x = 1.88168 E - 6

Solution steps

\frac{3 ^{2} x}{3 ^{- 7}} = 3^{- 3}

Simplify

3^{- 3} : \frac{1}{3 ^{3}}

\frac{3 ^{2} x}{3 ^{- 7}} = \frac{1}{3 ^{3}}

Multiply both sides by

3^{- 7}

3^{2} x = \frac{1}{3 ^{10}}

Divide both sides by

3^{2}

x = \frac{1}{3 ^{12}}

x, 2 x + y + 3 z = 1

\frac{x}{2} - \frac{( 3 x )}{4} + \frac{x}{6} = - \frac{2}{3}

x, f (x - 1) = - f (x + 1)

8 - x = 9

x - 1 + x - 2 + x - 3 = 6

What is the answer to (3^2x)/(3^{-7)}=3^{-3} ?
The answer to (3^2x)/(3^{-7)}=3^{-3} is x= 1/(3^{12)}
Solve for x: (3^2x)/(3^{-7)}=3^{-3}
The solution is x= 1/(3^{12)}