What is the answer to 64^{2x-5}=sqrt(2^{3-7x)} ?

The answer to 64^{2x-5}=sqrt(2^{3-7x)} is x= 63/31

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64^{2x-5}=sqrt(2^{3-7x)}

6 4^{2 x - 5} = 2^{3 - 7 x}

x = \frac{63}{31}

Decimal Notation

x = 2.03225 \dots

Solution steps

6 4^{2 x - 5} = 2^{3 - 7 x}

Apply exponent rules

6 (2 x - 5) = (3 - 7 x) \frac{1}{2}

Solve

6 (2 x - 5) = (3 - 7 x) \frac{1}{2} : x = \frac{63}{31}

x = \frac{63}{31}

3^{2 x} = 9^{- 2 x - 18}

9^{x - 4} = 7^{- 3 x}

3^{x} + 3^{x + 2} = \frac{10}{3}

30 x^{4} + 4 e^{2 x} = 0

x, e^{2 x + 1} - 3 = 0

What is the answer to 64^{2x-5}=sqrt(2^{3-7x)} ?
The answer to 64^{2x-5}=sqrt(2^{3-7x)} is x= 63/31
Solve for x: 64^{2x-5}=sqrt(2^{3-7x)}
The solution is x= 63/31