What is the answer to 3^{2-x}=(27^{x+1})/9 ?

The answer to 3^{2-x}=(27^{x+1})/9 is x= 1/4

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3^{2-x}=(27^{x+1})/9

3^{2 - x} = \frac{2 7 ^{x + 1}}{9}

x = \frac{1}{4}

Decimal Notation

x = 0.25

Solution steps

3^{2 - x} = \frac{2 7 ^{x + 1}}{9}

Multiply both sides by

9

3^{2 - x} \cdot 9 = \frac{2 7 ^{x + 1}}{9} \cdot 9

Simplify

3^{2 - x} \cdot 9 = 2 7^{x + 1}

Apply exponent rules

- x + 4 = 3 (x + 1)

Solve

- x + 4 = 3 (x + 1) : x = \frac{1}{4}

x = \frac{1}{4}

3^{\frac{x ^{2} - 4 x - 5}{3 x - 1}} = 1

256 = (\frac{1}{4})^{3 x + 2}

t, x = e^{4 t}

2^{y} = \frac{1}{32}

170 = 180 (1 - e^{- 0.5 t})

What is the answer to 3^{2-x}=(27^{x+1})/9 ?
The answer to 3^{2-x}=(27^{x+1})/9 is x= 1/4
Solve for x: 3^{2-x}=(27^{x+1})/9
The solution is x= 1/4