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Sagot :
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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