What is 7<= log_{2}(x)<= 8 ?

The solution to 7<= log_{2}(x)<= 8 is 128<= x<= 256

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7<= log_{2}(x)<= 8

7 \leq lo g_{2} (x) \leq 8

128 \leq x \leq 256

Solution steps

7 \leq lo g_{2} (x) \leq 8

a \leq u \leq b

then

a \leq u and u \leq b

7 \leq lo g_{2} (x) and lo g_{2} (x) \leq 8

7 \leq lo g_{2} (x) : x \geq 128

lo g_{2} (x) \leq 8 : 0 < x \leq 256

Combine the intervals

x \geq 128 and 0 < x \leq 256

Merge Overlapping Intervals

128 \leq x \leq 256

(x - 3)^{2} = 3 (x - 3)

(2 x - 1) (x + 1) = 6

1000 - \frac{6250}{313} + c = 50

\frac{2 Q ^{3} P}{Q P ^{- 3}}

250 x^{3} + 128 y^{3}

What is 7<= log_{2}(x)<= 8 ?
The solution to 7<= log_{2}(x)<= 8 is 128<= x<= 256