algebra
2007-12-31 18:23:42 · 6 answers · asked by Tim . 2 in Science & Mathematics ➔ Mathematics
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.
2007-12-31 18:26:46 · answer #1 · answered by johnson88 3 · 1⤊ 1⤋
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.
2016-10-10 18:47:11 · answer #2 · answered by ? 3 · 0⤊ 0⤋
2^5 = 32
Thus 2^x >32---->x > 5
2007-12-31 20:25:38 · answer #3 · answered by Como 7 · 2⤊ 0⤋
2^x > 32
2^x > 2^5
therefore , x > 5
2007-12-31 18:44:10 · answer #4 · answered by ck c 1 · 2⤊ 1⤋
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
2007-12-31 18:38:16 · answer #5 · answered by Puggy 7 · 2⤊ 4⤋
2 x2x2x2x2 = 32
therefore
X = 5
2007-12-31 18:29:17 · answer #6 · answered by jamesyoy02 6 · 0⤊ 5⤋