If $2^{2n-1} = \frac{1}{8^{n-3}}$, then the value of n is.

Explanation

(443) ## Page 42 $$2^{2n-1} = \frac{1}{8^{n-3}}$$ $$2^{2n-1} = \frac{1}{\left((2)^3\right)^{n-3}} = \frac{1}{2^{3n-9}}$$ $\therefore$ $2n - 1 = -(3n - 9)$ $2n - 1 = -3n + 9$ $5n = 10$ $n = 2$