Solve by factorization $4^{x + 1} + 4^{1 - x} = 10$

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Solution

The given equation is $4^{x + 1} + 4^{1 - x} = 10$
$4^{x + 1} + 4^{1 - x} = 10$
$\Rightarrow 4^{x} .4 + {4.4}^{- x} = 10 \Rightarrow 4. 4^{x} + \frac{4}{4^{x}} = 10 \Rightarrow 4. (4^{x})^{2} + 4 = {10.4}^{x} \Rightarrow 4. (4^{x})^{2} - {10.4}^{x} + 4 = 0$
Put $4^{x} = z ∴ 4 z^{2} - 10 z + 4 = 0 \Rightarrow 4 z^{2} - 8 z - 2 z + 4 = 0 \Rightarrow 4 z (z - 2) - 2 (z - 2) = 0 \Rightarrow (4 z - 2) (z - 2) = 0 \Rightarrow 4 z - 2 = 0 o r z - 2 = 0 \Rightarrow z = \frac{1}{2} o r z = 2 ∴ 4^{x} = \frac{1}{2} o r 4^{x} = 2 \Rightarrow 2^{2 x} = 2^{- 1} o r 2^{2 x} = 2^{1} \Rightarrow 2 x = - 1 o r 2 x = 1 \Rightarrow x = - \frac{1}{2} o r x = \frac{1}{2}$
Thus, the values of x are $- \frac{1}{2}$ or $\frac{1}{2}$

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