Zero and Negative Exponents?

Question

Evaluate without using technology

a)(6^0+3^-1)^-1
b)(2^-1-2^-3)^2
c) (2-a^-1)^2 <---see example below
d)(x^-2+y^-4)^-2

example (c) - 4a^2-4a+1/a^2

Answer 1

any number (except zero) to the zero power equals 1

Negative exponents in the numerator are positive in the denominator

Quotient to a negative power is the reciprocal to a positive power

So 6^0 = 1
3^-1 = 1/3
so 1 + 1/3 = 4/3
and (4/3)^1 = 3/4

2^-1 = 1/2
2^-3 = 1/8
so 1/2 - 1/8 = 3/8
and (3/8)^2 = 9/64

a^-1 = 1/a
2 - 1/a = (2a - 1)/a
so (2a - 1)/a squared =
[4a^2 - 4a + 1] / a^2

x^-2 = 1/x^2
y^-4 = 1/y^4
and 1/x^2 + 1/y^4 = [(y^4 + x^2) / (x^2y^4)]
[(y^4 + x^2) / (x^2y^4)]^-2 =
[(x^2y^4) / (y^4 + x^2)] ^2

Answer 2

Recall, anything to the 0 power is 1 so for (a)
(6^0+3^-1)^-1 = (1 + 3^-1)^-1
Now recall that x^(-a) = 1/(x^a). Now you get...
= (1 + 1/(3^1))^(-1) = (1 + 1/3)^(-1)
Simplify the sum to an improper fraction to get
= (4/3)^(-1)
Use the rule again to get
= 1/[(4/3)^1] = 1/(4/3)
Use division of fractions to get (using % as the division symbol)
= 1 % 4/3 = 1 * 3/4 = 3/4

For (b) same idea.
(2^(-1) - 2^(-3))^2 = (1/(2^1) - 1/(2^3))^2
= (1/2 - 1/8)^2
= (4/8 - 1/8)^2
= (3/8)^2
= (3^2)/(8^2)
= 9/64

For (c)
(2 - a^(-1))^2 = (2 - 1/(a^1))^2
= (2 - 1/a)^2
Now FOIL
= 4 - 2(2)(1/a) + (1/a)^2 = 4 - 4/a + 1/(a^2)

For (d)
[x^(-2) + y^(-4)]^(-2) = [1/(x^2) + 1/(y^4)]^(-2)
= [y^4/(x^2*y^4) + x^2/(x^2*y^4)]^(-2)
= [(y^4 + x^2)/(x^2*y^4)]^(-2)
= 1/[(y^4 + x^2)/(x^2*y^4)]^2
= [(x^2*y^4)^2]/[(y^4 + x^2)^2]
= (x^4*y^8)/[(y^4 + x^2)^2]
You can FOIL out the bottom, but I think that is sufficient. Hope this helps.

Answer 3

Anything with zero as the exponent is going to be 1, no matter what.

With negative exponents, you have to switch the number from a numerator to a denominator (or the other way around) to get a positive exponent. For example, if you have two to the negative third power, 2 ^ -3, you would make it into one over two to the third power, 1 / (2^3), so the answer would be 1/8.

Answer 4

a: (6⁰+3⁻¹)⁻¹

First resolve the stuff in the parentheses, then apply the exponent:

(1+1/3)⁻¹
(4/3)⁻¹
3/4

b: (2⁻¹-2⁻³)²

Same strategy as with the first one:

(1/2 - 1/8)²
(3/8)²
9/64

c: (2-a⁻¹)²

Since we have variables in the parentheses, we cannot resolve before applying the exponent, so instead we expand using the binomial theorem:

2²-2*2a⁻¹+a⁻²

Simplifying:

4-4/a+1/a²

This is equivalent to:

(4a²-4a+1)/a²

So I assume that is what you meant to put in your example.

d: (x⁻²+y⁻⁴)⁻²

First, expand the inside using the binomial theorem:

1/(x⁻⁴ + 2x⁻²y⁻⁴ + y⁻⁸)

Find a common denominator:

1/((y⁸ + 2x²y⁴ + x⁴)/(x⁴y⁸))

Simplify:

x⁴y⁸/(y⁸ + 2x²y⁴ + x⁴)

Answer 5

for any x and any a, x^0 = 1 and x^-a = 1/x^a

thus, a) 1/(1+1/3) = 3/4
b) (1/2 - 1/8)^2 = 9/64
c) (2- 1/a)^2 = ((2a-1)/a)^2 hard to simplify much...
d) again you can shape this one up a little bit, but youre not gonna get a real clean answer