# How do you solve this? 4^x=2^(x-1)-8

## Thanks!

Jun 26, 2018

$x \notin \mathbb{R}$

#### Explanation:

${4}^{x} = {2}^{x - 1} - 8$
${\left({2}^{2}\right)}^{x} = {2}^{x - 1} - 8$
${2}^{2 x} = \frac{1}{2} \cdot {2}^{x} - 8$
Let $X = {2}^{x}$
X²=1/2X-8
X²-1/2X+8
$\Delta = \frac{1}{4} - 32$
$\Delta = - \frac{127}{4}$
$\Delta < 0 \implies x \notin \mathbb{R}$

Jun 26, 2018

No real solutions

#### Explanation:

${\left({2}^{2}\right)}^{x} = {2}^{x - 1} - 8$

$\implies {2}^{2 x} - {2}^{x - 1} + 8 = 0$

$\implies {2}^{2 x} - \left({2}^{- 1} \cdot \left({2}^{x}\right)\right) + 8 = 0$

$\implies {\left({2}^{x}\right)}^{2} - \frac{1}{2} \left({2}^{x}\right) + 8 = 0$

Let $l a m \mathrm{da} = {2}^{x}$

$\implies l a m {\mathrm{da}}^{2} - \frac{1}{2} l a m \mathrm{da} + 8 = 0$

$\implies \text{ discriminant " < 0 -> "no solutions for } l a m \mathrm{da}$

Hence no solutions for $l a m \mathrm{da}$ in real, means no solutions for ${2}^{x}$ hence no solutions for $x$

Jun 26, 2018

Continuing:

If you understand complex numbers

#### Explanation:

$l a m \mathrm{da} = \frac{1}{4} \pm \frac{\sqrt{127}}{4} i$

lamda = 2sqrt2 e^(pmiarctan(sqrt(127) )

 = 2sqrt2 e^(pmiarctan(sqrt(127) )) e^(2kpii

as ${e}^{2 k \pi i} = 1 , \forall k \in \mathbb{Z}$

=> e^(xln2) = 2sqrt2 e^(i (pmarctan(sqrt(127) )+ 2kpi)

$\implies x \ln 2 = \frac{3}{2} \ln 2 + i \left(2 k \pi \pm \arctan \sqrt{127}\right)$

$\implies x = \frac{3}{2} + \frac{i \left(2 k \pi \pm \arctan \sqrt{127}\right)}{\ln} 2$

$\forall k \in \mathbb{Z}$