What is the answer to 81^{x-3}=(1/9)^{2x-1} ?

The answer to 81^{x-3}=(1/9)^{2x-1} is x= 7/4

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

8 1^{x - 3} = (\frac{1}{9})^{2 x - 1}

x = \frac{7}{4}

Solution steps

8 1^{x - 3} = (\frac{1}{9})^{2 x - 1}

Apply exponent rules

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

Solve

2 (x - 3) = - (2 x - 1) : x = \frac{7}{4}

x = \frac{7}{4}

\frac{1}{25} = 5^{x + 4}

7^{x - 10} = 25

8^{(x^{2} + 2 x - 15)} = 1

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

9^{x} = 1

What is the answer to 81^{x-3}=(1/9)^{2x-1} ?
The answer to 81^{x-3}=(1/9)^{2x-1} is x= 7/4
Solve for x: 81^{x-3}=(1/9)^{2x-1}
The solution is x= 7/4