2^8/8^x=4^x

Solution for 2^8/8^x=4^x equation:

2^8/8^x=4^x

We move all terms to the left:

2^8/8^x-(4^x)=0


Domain of the equation: 8^x!=0

x!=0/1

x!=0

x∈R


We multiply all the terms by the denominator


-4^x*8^x+2^8=0

We add all the numbers together, and all the variables

-4^x*8^x+256=0

Wy multiply elements

-32x^2+256=0

a = -32; b = 0; c = +256;
Δ = b2-4ac
Δ = 02-4·(-32)·256
Δ = 32768
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{32768}=\sqrt{16384*2}=\sqrt{16384}*\sqrt{2}=128\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-128\sqrt{2}}{2*-32}=\frac{0-128\sqrt{2}}{-64}
=-\frac{128\sqrt{2}}{-64}
=-\frac{2\sqrt{2}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+128\sqrt{2}}{2*-32}=\frac{0+128\sqrt{2}}{-64}
=\frac{128\sqrt{2}}{-64}
=\frac{2\sqrt{2}}{-1}
$