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![{4}^{x} - 5 \times {2}^{x} + 4 = 0 {4}^{x} - 5 \times {2}^{x} + 4 = 0](https://tex.z-dn.net/?f=+%7B4%7D%5E%7Bx%7D++-+5++%5Ctimes++%7B2%7D%5E%7Bx%7D++%2B+4+%3D+0)
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11
Answer:
0 or 2
Step-by-step explanation:
Given :
4ˣ - 5 × 2ˣ + 4 = 0
2ˣ × 2ˣ - 5 × 2ˣ + 4 = 0
Let 2ˣ = y
y² - 5 y + 4 = 0
y² - 4 y - y + 4 = 0
y ( y - 4 ) - ( y - 4 ) = 0
( y - 1 ) ( y - 4 ) = 0
y - 1 = 0
y = 1
OR
y - 4 = 0
y = 4
Now on comparing :
2ˣ = 4
2ˣ = 2²
x = 2
OR
2ˣ = 1
2ˣ = 2⁰
x = 0
Therefore , the value of x is 0 OR 2 .
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