Question 77

If $$2^{2n - 1} = \frac{1}{8^{n - 3}}$$, then $$2^{n - 2} =$$

Solution

Given, $$2^{2n-1}=\frac{1}{8^{n-3}}$$

$$\Rightarrow$$ $$2^{2n-1}=\frac{1}{2^{3\left(n-3\right)}}$$

$$\Rightarrow$$ $$2^{2n-1}=2^{-3\left(n-3\right)}$$

As bases are equal, equating powers

$$2n-1=-3\left(n-3\right)$$

$$\Rightarrow$$ $$2n-1=-3n+9$$

$$\Rightarrow$$ $$5n=10$$

$$\Rightarrow$$ $$n=2$$

$$\therefore\ $$ $$2^{n-2}=2^{2-2}=2^0=1$$

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