If `2^(2x-1) = 1/(8^(x-3))`, then the value of `x` is:
\[ \text { If } 2^{2 x-1}=\frac{1}{8^{x-3}} \text {, then the value of } x \text { is } \]
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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.