solve 2^x >32?

Question

algebra

Answer 1

x > 5

2 ^ 5 = 32, therefore x can never equal 5.

This equation wants to know what is the value of x such that 2 to some power of x is GREATER than 32, not equal to 32, not less than 32. So, if 2 ^ 5 = 32, then the value of x must be at least GREATER than 5.

Answer 2

2x^2 > 32 ? divide the two aspects via 2 x^2 > sixteen ? take the sq. root of the two aspects x > ±4 x should be extra desirable than 4 or decrease than a unfavorable 4. stable success on your learn, ~ Mitch ~ Edit: Forgot to divide the 32 via 2 and gave the incorrect answer. .... my undesirable.

Answer 3

2^5 = 32
Thus 2^x >32---->x > 5

Answer 4

2^x > 32
2^x > 2^5
therefore , x > 5

Answer 5

2^x > 32

Take the log[base 2] of both sides. Since log[base 2](x) is a function which is always increasing, it follows that if a > b, then log[base 2](a) > log[base 2](b).

log[base 2](2^x) > log[base 2](32)

Using the various log identities and manipulation,

x log[base 2](2) > log[base 2](2^5)

x (1) > 5 log[base 2](2)

x > 5(1)

x > 5

Answer 6

2 x2x2x2x2 = 32
therefore
X = 5