What is 2^{2x}-5(2^x)+4<0 ?

The solution to 2^{2x}-5(2^x)+4<0 is 0<x<2

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2^{2x}-5(2^x)+4<0

2^{2 x} - 5 (2^{x}) + 4 < 0

0 < x < 2

Solution steps

2^{2 x} - 5 (2^{x}) + 4 < 0

Apply exponent rules

(2^{x})^{2} - 5 \cdot 2^{x} + 4 < 0

Factor

(2^{x})^{2} - 5 \cdot 2^{x} + 4 : (2^{x} - 1) (2^{x} - 4)

(2^{x} - 1) (2^{x} - 4) < 0

Find the signs of the factors of

(2^{x} - 1) (2^{x} - 4)

Identify the intervals:

x < 0, x = 0, 0 < x < 2, x = 2, x > 2

Summarize in a table:

2^{x} - 1 2^{x} - 4 (2^{x} - 1) (2^{x} - 4) x < 0 - - + x = 0 0 - 0 0 < x < 2 + - - x = 2 + 00 x > 2 + + +

Identify the intervals that satisfy the required condition:

< 0

0 < x < 2

5 x - 2 = 4 x

9 x^{2} - 60 x + 100

2 x^{2} - 5 x - 3 < 0

9 y^{2} - 81

3 x + 5 \leq 14