What is the answer to \sqrt[3x-1]{9}=9^x ?

The answer to \sqrt[3x-1]{9}=9^x is x=(1+sqrt(13))/6 ,x=(1-sqrt(13))/6

Solve for x: \sqrt[3x-1]{9}=9^x

The solution is x=(1+sqrt(13))/6 ,x=(1-sqrt(13))/6

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x = \frac{1 + 13}{6}, x = \frac{1 - 13}{6}

Decimal Notation

x = 0.76759 \dots, x = - 0.43425 \dots

Solution steps

Apply exponent rules

\frac{1}{3 x - 1} = x

Solve

\frac{1}{3 x - 1} = x : x = \frac{1 + 13}{6}, x = \frac{1 - 13}{6}

x = \frac{1 + 13}{6}, x = \frac{1 - 13}{6}

3 e^{3 x} - e^{x} = 0

2 n = 4^{p} \cdot 8^{n}

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

2 (2^{2 x}) - 6 (2^{x}) + 4 = 0

12 = 200 (1 - e^{- \frac{t}{800}})

What is the answer to \sqrt[3x-1]{9}=9^x ?
The answer to \sqrt[3x-1]{9}=9^x is x=(1+sqrt(13))/6 ,x=(1-sqrt(13))/6
Solve for x: \sqrt[3x-1]{9}=9^x
The solution is x=(1+sqrt(13))/6 ,x=(1-sqrt(13))/6