Question

Solve for x not y please I am stuck

Answer 1

Answer:

x=2, x=1

Step-by-step explanation:

Answer 2

let's solve :

• ${4}^{2x} - 20(4 {}^{x} ) + 64 = 0$

let's take

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

now,

• ${(2{}^{2x} )} {}^{2} - 20(2) {}^{2x} + 64 = 0$

• $y {}^{2} - 20y + 64 = 0$

• $y {}^{2} - 16y - 4y + 64 = 0$

• $y(y - 16) - 4( - 16) = 0$

• $(y - 16) (y - 4) = 0$

• $y = 16 \: \: or \: \: y = 4$

if y = 16, then :

• $2 {}^{2x} = 16$

• ${2}^{2x} = 2 {}^{4}$

• $2x = 4$

• $x = 2$

and if y = 4 :

• ${2}^{2x} = 4$

• ${2}^{2x} = {2}^{2}$

• $2x = 2$

• [tex]x = 1[/tex

hence, value of x = 1 or x = 2