What is the answer to 128\times 16^{2x+1}=8^{3-2x} ?

The answer to 128\times 16^{2x+1}=8^{3-2x} is x=-1/7

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128\times 16^{2x+1}=8^{3-2x}

128 \cdot 1 6^{2 x + 1} = 8^{3 - 2 x}

x = - \frac{1}{7}

Decimal Notation

x = - 0.14285 \dots

Solution steps

128 \cdot 1 6^{2 x + 1} = 8^{3 - 2 x}

Apply exponent rules

4 (2 x + 1) + 7 = 3 (3 - 2 x)

Solve

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

x = - \frac{1}{7}

y, x^{x} = y^{y}

(\frac{1}{2})^{x} = 128

5 e^{x} - 12 e^{- x} - 11 = 0

7^{3 x - 1} = 5^{x + 1}

2^{2 x - 4} = 8^{4}

What is the answer to 128\times 16^{2x+1}=8^{3-2x} ?
The answer to 128\times 16^{2x+1}=8^{3-2x} is x=-1/7
Solve for x: 128\times 16^{2x+1}=8^{3-2x}
The solution is x=-1/7