Sunsha wrote:
If \((-\frac{1}{2})^N >-8\), which of the following could be the value of \(N\)?
Indicate all that apply
A -10
B -7
C -3
D 0
E 3
F 10
Two important rules:
ODD exponents preserve the sign of the base. So, (
NEGATIVE)^(
ODD integer) =
NEGATIVEand (
POSITIVE)^(
ODD integer) =
POSITIVEAn EVEN exponent always yields a positive result (unless the base = 0)
So, (
NEGATIVE)^(
EVEN integer) =
POSITIVEand (
POSITIVE)^(
EVEN integer) =
POSITIVESince \((-\frac{1}{2})^N \) will be positive for all EVEN values of N, we can see that N = -10, 0 and 10, will yield powers that are greater than -8
So, answer choices A, D and F work so far.
Let's just test the remaining answer choices...
B) -7Useful rule: \((\frac{a}{b})^{-k} = (\frac{b}{a})^{k}\)So, \((-\frac{1}{2})^{-7} = (-\frac{2}{1})^{7}=(-2)^{7}=-128\)
Since \(-128 < -8\), answer choice B isn't a solution.
C) -3\((-\frac{1}{2})^{-3} = (-\frac{2}{1})^{3}=(-2)^{3}=-8\)
Since \(-8 = -8\), answer choice C isn't a solution.
E) 3\((-\frac{1}{2})^{3} = -\frac{1}{8}\)
Since \(-\frac{1}{8} > -8\), answer choice E works.
Answer: A, D, E, F