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

The answer to 27^{2t+1}=(1/9)^{t-1} is t=-1/8

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

2 7^{2 t + 1} = (\frac{1}{9})^{t - 1}

t = - \frac{1}{8}

Decimal Notation

t = - 0.125

Solution steps

2 7^{2 t + 1} = (\frac{1}{9})^{t - 1}

Apply exponent rules

3 (2 t + 1) = - 2 (t - 1)

Solve

3 (2 t + 1) = - 2 (t - 1) : t = - \frac{1}{8}

t = - \frac{1}{8}

5^{x + 2} = 4^{2 x - 1}

5^{x + 2} = 4^{2 x - 2}

e^{e} - 12 e^{- x} - 1 = 0

5 = 10 (0.8)^{x}

2 7^{x} = 9^{x - 4}

What is the answer to 27^{2t+1}=(1/9)^{t-1} ?
The answer to 27^{2t+1}=(1/9)^{t-1} is t=-1/8
Solve for t: 27^{2t+1}=(1/9)^{t-1}
The solution is t=-1/8