Answer:
Here we will use the relationships:
[tex]a^n*a^m = a^{n + m}[/tex]
[tex](a^n)^m = a^{n*m}[/tex]
[tex]\frac{a^n}{a^m} = a^{n - m}[/tex]
And a number:
[tex]a^n[/tex]
is between 0 and 1 if a is positive and larger than 1, and n is negative.
if a is positive and 0 < a < 1, then we need to have n positive such that:
0 < a^n < 1
A) [tex]5^3*5^{-4} = 5^{3 + (-4)} = 5^{-1}[/tex]
This is between zero and 1,
B) [tex]\frac{3^5}{3^{-6}} = 3^{5 - (-6)} = 3^{11}[/tex]
This is greater than 1, because the exponent is positive.
C) [tex](\frac{1}{4})^3*( \frac{1}{4})^2 = (\frac{1}{4})^{2 + 3} = (\frac{1}{4})^5[/tex]
Because a is smaller than 1, and the exponent is positive, then the expression is between 0 and 1.
D) [tex]\frac{(-7)^5}{(-7)^7} = (-7)^{5 - 7} = -7^{-2}[/tex]
The exponent is negative (and pair) then the expression is between 0 and 1.
Remember that when the exponent is pair, we always have that:
(-N)^m = (N)^m
So (-7)^-2 = 7^-2
