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Answer :

Answer:

x=2, x=1

Step-by-step explanation:

DᴀʀᴋPᴀʀᴀᴅᴏxgo

let's solve :

  • [tex] {4}^{2x} - 20(4 {}^{x} ) + 64 = 0[/tex]

let's take

[tex]2 {}^{2x} = y[/tex]

now,

  • [tex]{(2{}^{2x} )} {}^{2} - 20(2) {}^{2x} + 64 = 0[/tex]

  • [tex]y {}^{2} - 20y + 64 = 0[/tex]

  • [tex]y {}^{2} - 16y - 4y + 64 = 0[/tex]

  • [tex]y(y - 16) - 4( - 16) = 0[/tex]

  • [tex](y - 16) (y - 4) = 0[/tex]

  • [tex]y = 16 \: \: or \: \: y = 4[/tex]

if y = 16, then :

  • [tex]2 {}^{2x} = 16[/tex]

  • [tex] {2}^{2x} = 2 {}^{4} [/tex]

  • [tex]2x = 4[/tex]

  • [tex]x = 2[/tex]

and if y = 4 :

  • [tex] {2}^{2x} = 4[/tex]

  • [tex] {2}^{2x} = {2}^{2} [/tex]

  • [tex]2x = 2[/tex]

  • [tex]x = 1[/tex

hence, value of x = 1 or x = 2

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