If $3^{x + 1} = 9^{x - 2}, f i n d t h e v a l u e o f 2^{1 + x}$ .

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Solution

$3^{x + 1} = 9^{x - 2} \Rightarrow 3^{x + 1} = (3^{2})^{x - 2} \Rightarrow 3^{x + 1} = 3^{2 x - 4} ∴ x + 1 = 2 x - 4 \Rightarrow 1 + 4 = 2 x - x \Rightarrow x = 5 N o w, 2^{1 + x} = 2^{1 + 5} = 2^{6} = 64$

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