What is the answer to (1/9)^{3x-1}=27^{5x-4} ?

The answer to (1/9)^{3x-1}=27^{5x-4} is x= 2/3

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(1/9)^{3x-1}=27^{5x-4}

(\frac{1}{9})^{3 x - 1} = 2 7^{5 x - 4}

x = \frac{2}{3}

Decimal Notation

x = 0.66666 \dots

Solution steps

(\frac{1}{9})^{3 x - 1} = 2 7^{5 x - 4}

Apply exponent rules

- 2 (3 x - 1) = 3 (5 x - 4)

Solve

- 2 (3 x - 1) = 3 (5 x - 4) : x = \frac{2}{3}

x = \frac{2}{3}

2 e^{2 x} = e^{x}

2 e^{\frac{1}{x}} = 42

3 2^{- 2 k} = 64

j, z = e^{- \frac{π}{4} j}

3^{x} + 3^{- x} = 5

What is the answer to (1/9)^{3x-1}=27^{5x-4} ?
The answer to (1/9)^{3x-1}=27^{5x-4} is x= 2/3
Solve for x: (1/9)^{3x-1}=27^{5x-4}
The solution is x= 2/3