What is the answer to 8^{2x+1}=\sqrt[3]{4^{x-1}} ?

The answer to 8^{2x+1}=\sqrt[3]{4^{x-1}} is x=-11/16

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8^{2x+1}=\sqrt[3]{4^{x-1}}

x = - \frac{11}{16}

Decimal Notation

x = - 0.6875

Solution steps

Apply exponent rules

3 (2 x + 1) = \frac{2}{3} (x - 1)

Solve

3 (2 x + 1) = \frac{2}{3} (x - 1) : x = - \frac{11}{16}

x = - \frac{11}{16}

b^{2^{2 x - 17}} = b^{128}

3^{x - 1} + 3^{x} + 3^{x + 1} = 117

2^{2} n = 4^{n}

e^{k} = 2

e^{x} = e^{- 4 x}

What is the answer to 8^{2x+1}=\sqrt[3]{4^{x-1}} ?
The answer to 8^{2x+1}=\sqrt[3]{4^{x-1}} is x=-11/16
Solve for x: 8^{2x+1}=\sqrt[3]{4^{x-1}}
The solution is x=-11/16