\[ \text { If } 2^{2 x-1}=\frac{1}{8^{x-3}} \text {, then the value of } x \text { is } \]

We can rewrite the equation \(2^{2x – 1} = \frac{1}{8^{x – 3}}\) using the properties of exponents:

\[2^{2x – 1} = 2^{-3(x – 3)}\]

This is because \(8 = 2^3\), so \(8^{x – 3} = (2^3)^{x – 3} = 2^{3(x – 3)}\).

Since the bases are the same, we can set the exponents equal to each other:

\[2x – 1 = -3(x – 3)\]

Expanding:

\[2x – 1 = -3x + 9\]

Adding \(3x\) to both sides and adding 1 to both sides:

\[5x = 10\]

Dividing both sides by 5:

\[x = 2\]

Therefore, the value of \(x\) is 2.