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# How can this be solved??? help !!

## ${4}^{x + 1} - 9 \cdot {2}^{x} + 2 \ne 0$

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#### Explanation

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#### Explanation:

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1s2s2p Share
Feb 9, 2018

$x \ne - 2 \mathmr{and} 1$

#### Explanation:

We have ${4}^{x + 1} - 9 \cdot {2}^{x} + 2 \ne 0$

${4}^{x + 1} = {\left({2}^{x + 1}\right)}^{2} = {2}^{2 x + 2}$

Now we have:
$4 \left({2}^{2 x}\right) - 9 \left({2}^{x}\right) + 2 \ne 0$

Let ${2}^{x} = y$

$4 {y}^{2} - 9 y + 2 \ne 0$

$y = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$y = \frac{- \left(- 9\right) \pm \sqrt{{\left(- 9\right)}^{2} - 4 \left(4 \cdot 2\right)}}{2 \cdot 4}$

$y = \frac{9 \pm \sqrt{81 - 32}}{8}$

$y = \frac{9 \pm \sqrt{49}}{8}$

$y = \frac{9 \pm 7}{8}$

$y = \frac{1}{4} \mathmr{and} 2$

${2}^{x} = \frac{1}{4} \mathmr{and} 2$

$x = {\log}_{2} \left(\frac{1}{4}\right) \mathmr{and} {\log}_{2} \left(2\right)$

$x = - 2 \mathmr{and} 1$

So, for ${4}^{x + 1} - 9 \cdot {2}^{x} + 2 \ne 0$, $x \ne - 2 \mathmr{and} 1$

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