Negative Exponents Questions with Detailed Solutions for Grade 8
Detailed solutions and explanations to
Grade 8 questions on simplifying expressions with negative exponents .
Review: Rules of Exponents
- Rule 1: \( (a \times b)^n = a^n \times b^n \)
- Rule 2: \( \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \)
- Rule 3: \( a^{-n} = \frac{1}{a^n} \)
- Rule 4: \( \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n \)
-
Simplify the expression \(5^{-2}\)
Solution
Using the negative exponent rule:
\[
5^{-2} = \frac{1}{5^2} = \frac{1}{25}
\]
-
Simplify the expression \(\left(-\frac{1}{3}\right)^{-2}\)
Solution
Rewrite as a product:
\[
\left(-\frac{1}{3}\right)^{-2} = (-1)^{-2} \cdot \left(\frac{1}{3}\right)^{-2}
\]
Apply negative exponent rule:
\[
= \frac{1}{(-1)^2} \cdot \left(\frac{3}{1}\right)^2 = 1 \cdot 9 = 9
\]
-
Simplify \(\frac{5^{-1}}{3^{-1}}\)
Solution
Rewrite using quotient of powers:
\[
\frac{5^{-1}}{3^{-1}} = \left(\frac{5}{3}\right)^{-1} = \frac{3}{5}
\]
-
Simplify \((2^{-3})(3^{-2})\)
Solution
Rewrite using negative exponent rule:
\[
(2^{-3})(3^{-2}) = \frac{1}{2^3} \cdot \frac{1}{3^2} = \frac{1}{8} \cdot \frac{1}{9} = \frac{1}{72}
\]
-
Simplify \(-2^{-3}\)
Solution
Rewrite:
\[
-2^{-3} = (-1) \cdot 2^{-3} = (-1) \cdot \frac{1}{2^3} = -\frac{1}{8}
\]
-
Simplify \(-1^{-3} + 2^{-3}\)
Solution
Rewrite:
\[
-1^{-3} + 2^{-3} = (-1) \cdot \frac{1}{1^3} + \frac{1}{2^3} = -1 + \frac{1}{8}
\]
Find common denominator:
\[
= -\frac{8}{8} + \frac{1}{8} = -\frac{7}{8}
\]
-
Simplify \((-1)^{-4} + 2^{-3}\)
Solution
Rewrite:
\[
(-1)^{-4} + 2^{-3} = \frac{1}{(-1)^4} + \frac{1}{2^3} = 1 + \frac{1}{8}
\]
Find common denominator:
\[
= \frac{8}{8} + \frac{1}{8} = \frac{9}{8}
\]
-
Simplify \(\left(-4^{-2}\right)(2^2)\)
Solution
Rewrite:
\[
(-4^{-2})(2^2) = - \frac{1}{4^2} \cdot 2^2 = -\frac{1}{16} \cdot 4 = -\frac{1}{4}
\]
-
Simplify \((-1)^{-3} + 2^{0}\)
Solution
Note that \(2^0 = 1\). Rewrite:
\[
(-1)^{-3} + 1 = \frac{1}{(-1)^3} + 1 = -1 + 1 = 0
\]
-
Simplify \(0^{-3}\)
Solution
Rewrite:
\[
0^{-3} = \frac{1}{0^3} = \frac{1}{0}
\]
Division by zero is undefined, so \(0^{-3}\) is not a real number.
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