# Question #b397b

Aug 19, 2017

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

#### Explanation:

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

as

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

As you know, you have

$4 = {2}^{2}$

This implies that

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

This means that the equation can be written as

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

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

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

This is equivalent to

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

${2}^{2 x} = 2$

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

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

This implies that

$2 x = 1$

which gets you

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

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

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

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

Since

${4}^{\frac{3}{2}} = \sqrt{{4}^{3}} = 4 \sqrt{4} = 4 \cdot 2 = 8$

you will have

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

$2 - 40 + 38 = 0 \text{ } \textcolor{\mathrm{da} r k g r e e n}{\sqrt{}}$

Aug 19, 2017

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

#### Explanation:

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

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

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

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

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

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

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

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

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

Dividing by $- 19$:

${2}^{2 x} = 2$

${2}^{\textcolor{b l u e}{2 x}} = {2}^{\textcolor{b l u e}{1}}$

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

$2 x = 1$

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