Question #b397b

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Question #b397b

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Answer 1 · 2017-08-19T14:59:46

$x = \frac{1}{2}$ $x = 1/2$

Explanation:

Start by rewriting your equation

$2^{2 x} - 5 \cdot 4^{x + 1} + 38 = 0$ $2^(2x) - 5 * 4^(x+1) + 38 = 0$

as

$2^{2 x} - 5 \cdot 4^{x} \cdot 4 + 38 = 0$ $2^(2x) - 5 * 4^x * 4 + 38 = 0$

As you know, you have

$4 = 2^{2}$ $4 = 2^2$

This implies that

$4^{x} = {(2^{2})}^{x} = 2^{2 \cdot x} = 2^{2 x}$ $4^x = (2^2)^x = 2^(2 * x) = 2^(2x)$

This means that the equation can be written as

$2^{2 x} - 5 \cdot 4 \cdot 2^{2 x} + 38 = 0$ $2^(2x) - 5* 4 * 2^(2x) + 38 = 0$

At this point, you can take $2^{2 x}$ $2^(2x)$ as a common factor and say that

$2^{2 x} \cdot (1 - 5 \cdot 4) + 38 = 0$ $2^(2x) * (1 - 5 * 4) + 38 = 0$

This is equivalent to

$2^{2 x} = \frac{- 38}{- 19}$ $2^(2x) = (- 38)/(-19)$

$2^{2 x} = 2$ $2^(2x) = 2$

Since $2$ $2$ is simply $2^{1}$ $2^1$ , you can say that

$2^{2 x} = 2^{1}$ $2^(2x) = 2^1$

This implies that

$2 x = 1$ $2x = 1$

which gets you

$x = \frac{1}{2}$ $x = 1/2$

To double-check your calculations, plug $x = \frac{1}{2}$ $x = 1/2$ into the original equation.

$2^{(2 \cdot \frac{1}{2})} - 5 \cdot 4^{(\frac{1}{2} + 1)} + 38 = 0$ $2^((2 * 1/2)) - 5 * 4^((1/2 + 1)) + 38 = 0$

$2^{1} - 5 \cdot 4^{\frac{3}{2}} + 38 = 0$ $2^1 - 5 * 4^(3/2) + 38 = 0$

Since

$4^{\frac{3}{2}} = \sqrt{4^{3}} = 4 \sqrt{4} = 4 \cdot 2 = 8$ $4^(3/2) = sqrt(4^3) = 4sqrt(4) = 4 * 2 = 8$

you will have

$2 - 5 \cdot 8 + 38 = 0$ $2 - 5 * 8 + 38 = 0$

$2 - 40 + 38 = 0 \sqrt{}$ $2 - 40 + 38 = 0 " "color(darkgreen)(sqrt())$

Answer 2 · 2017-08-19T15:00:34

$x = \frac{1}{2}$ $x=1/2$

Explanation:

Note first that $5 (4^{x + 1}) = 5 (4^{x}) 4^{1} = 20 (4^{x})$ $5(4^(x+1))=5(4^x)4^1=20(4^x)$ .

$2^{2 x} - 5 (4^{x + 1}) + 38 = 0$ $2^(2x)-5(4^(x+1))+38=0$

$2^{2 x} - 20 (4^{x}) + 38 = 0$ $2^(2x)-20(4^x)+38=0$

Rewrite the exponential function with base $4$ $4$ as one with base $2$ $2$ so that we are working with a standard base throughout. Note that $4^{x} = {(2^{2})}^{x} = 2^{2 x}$ $4^x=(2^2)^x=2^(2x)$ .

$2^{2 x} - 20 (2^{2 x}) + 38 = 0$ $2^(2x)-20(2^(2x))+38=0$

Now, note that $2^{2 x} - 20 (2^{2 x}) = - 19 (2^{2 x})$ $2^(2x)-20(2^(2x))=-19(2^(2x))$ . This is just like how $x - 20 x = - 19 x$ $x-20x=-19x$ .

$- 19 (2^{2 x}) + 38 = 0$ $-19(2^(2x))+38=0$

To solve for $x$ $x$ , first isolate $2^{2 x}$ $2^(2x)$ .

$- 19 (2^{2 x}) = - 38$ $-19(2^(2x))=-38$

Dividing by $- 19$ $-19$ :

$2^{2 x} = 2$ $2^(2x)=2$

$2^{2 x} = 2^{1}$ $2^color(blue)(2x)=2^color(blue)1$

Since the bases of the exponential functions are equal, so must their exponents:

$2 x = 1$ $2x=1$

$x = \frac{1}{2}$ $x=1/2$

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