minus 3 by 2 to the power minus 1 into 2 by 3 to the power 2 divided by 2 by 3 to the power minus 1 + 3 by 2
Answers
Step-by-step explanation:
We define a negative power as the multiplicative inverse of the base raised to the positive opposite of the power:
x^{-n}=\dfrac{1}{x^n}x
−n
=
x
n
1
x, start superscript, minus, n, end superscript, equals, start fraction, 1, divided by, x, start superscript, n, end superscript, end fraction
Want to learn more about this definition? Check out this video.
Examples
3^{-5}=\dfrac{1}{3^5}3
−5
=
3
5
1
3, start superscript, minus, 5, end superscript, equals, start fraction, 1, divided by, 3, start superscript, 5, end superscript, end fraction
\dfrac{1}{2^8}=2^{-8}
2
8
1
=2
−8
start fraction, 1, divided by, 2, start superscript, 8, end superscript, end fraction, equals, 2, start superscript, minus, 8, end superscript
y^{-2}=\dfrac{1}{y^{2}}y
−2
=
y
2
1
y, start superscript, minus, 2, end superscript, equals, start fraction, 1, divided by, y, squared, end fraction
\left(\dfrac{8}{6}\right)^{-3}=\left(\dfrac{6}{8}\right)^{3}(
6
8
)
−3
=(
8
6
)
3
left parenthesis, start fraction, 8, divided by, 6, end fraction, right parenthesis, start superscript, minus, 3, end superscript, equals, left parenthesis, start fraction, 6, divided by, 8, end fraction, right parenthesis, cubed
Practice
PROBLEM 1
Select the equivalent expression.
4^{-3}=?4
−3
=?4, start superscript, minus, 3, end superscript, equals, question mark
Choose 1 answer:
Choose 1 answer:
(Choice A)
A
\dfrac{1}{4^{3}}
4
3
1
start fraction, 1, divided by, 4, cubed, end fraction
(Choice B)
B
\dfrac{1^3}{4}
4
1
3
start fraction, 1, cubed, divided by, 4, end fraction
(Choice C)
C
-4^{3}−4
3
minus, 4, cubed
Explain
Want to try more problems like these? Check out this exercise.
Some intuition
So why do we define negative exponents this way? Here are a couple of justifications:
Justification #1: Patterns
nnn 2^n2
n
2, start superscript, n, end superscript
333 2^3=82
3
=82, cubed, equals, 8
222 2^2=42
2
=42, squared, equals, 4
111 2^1=22
1
=22, start superscript, 1, end superscript, equals, 2
000 2^0=12
0
=12, start superscript, 0, end superscript, equals, 1
-1−1minus, 1 2^{-1}=\dfrac122
−1
=
2
1
2, start superscript, minus, 1, end superscript, equals, start fraction, 1, divided by, 2, end fraction
-2−2minus, 2 2^{-2}=\dfrac142
−2
=
4
1
2, start superscript, minus, 2, end superscript, equals, start fraction, 1, divided by, 4, end fraction
Notice how 2^n2
n
2, start superscript, n, end superscript is divided by 222 each time we reduce nnn. This pattern continues even when nnn is zero or negative.
Justification #2: Exponent properties
Recall that \dfrac{x^n}{x^m}=x^{n-m}
x
m
x
n
=x
n−m
start fraction, x, start superscript, n, end superscript, divided by, x, start superscript, m, end superscript, end fraction, equals, x, start superscript, n, minus, m, end superscript. So...
\begin{aligned} \dfrac{2^2}{2^3}&=2^{2-3} \\\\ &=2^{-1} \end{aligned}
2
3
2
2
=2
2−3
=2
−1
We also know that
\begin{aligned} \dfrac{2^2}{2^3}&=\dfrac{\cancel 2\cdot\cancel 2}{\cancel 2\cdot\cancel 2\cdot 2} \\\\ &=\dfrac12 \end{aligned}
2
3
2
2
=
2
⋅
2
⋅2
2
⋅
2
=
2
1
And so we get 2^{-1}=\dfrac122
−1
=
2
1
2, start superscript, minus, 1, end superscript, equals, start fraction, 1, divided by, 2, end fraction.
Also, recall that x^n\cdot x^m=x^{n+m}x
n
⋅x
m
=x
n+m
x, start superscript, n, end superscript, dot, x, start superscript, m, end superscript, equals, x, start superscript, n, plus, m, end superscript. So...
\begin{aligned} 2^2\cdot 2^{-2}&=2^{2+(-2)} \\\\ &=2^0 \\\\ &=1 \end{aligned}
2
2
⋅2
−2
=2
2+(−2)
=2
0
=1
And indeed, according to the definition...
\begin{aligned} 2^2\cdot 2^{-2}&=2^2\cdot\dfrac{1}{2^2} \\\\ &=\dfrac{2^2}{2^2} \\\\ &=1 \end{aligned}
2
2
⋅2
−2
=2
2
⋅
2
2
1
=
2
2
2
2
=1
Answer:
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to
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