Index Laws
Recall, the following laws for simplifying indices.
… 1 … $a^m \times a^n = a^{m+n}$
… 2 … $\dfrac{a^m}{a^n} = a^{m - n}$
… 3 … $\big( a^m \big)^n = a^{mn}$
… 4 … $a^0 = 1$
… 5 … $a^{-m} = \dfrac{1}{a^m}$
… 6 … $a^ {\frac{1}{n}} = \sqrt[n]{a}$
… 7 … $a^{\frac{m}{n}} =\sqrt[n]{a^m} = \big( \sqrt[n]{a} \big)^m$
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Example 1
Simplify $15(x^2y^{-2})^4 \div \left( 3(x^4y)^{-2} \right)$
Solution:
… … $15 \big( x^2y^{-2} \big)^4 \div \left( 3 \big( x^4y \big)^{-2} \right)$
… … $= \dfrac{15x^8 y^{-8}} {3x^{-8} y^{-2}}$
… … $=5x^{16}y^{-6}$
… … $=\dfrac{5x^{16}}{y^6}$
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Example 2
Find $125^{\frac{2}{3}}$
Solution:
… … $125^{\frac{2}{3}}$
… … $= \big( \sqrt[3]{125} \big)^2$
… … $=5^2$
… … $=25$
Alternate Solution
… … $125^{\frac{2}{3}}$
… … $=\big(5^3\big)^\frac{2}{3}$
… … $=5^{(3\times\frac{2}{3})}$
… … $=5^2$
… … $=25$
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Example 3
Find $81^{-\frac{1}{4}}$
Solution:
… … $81^{-\frac{1}{4}}$
… … $= \big( 3^4 \big)^{-\frac{1}{4}}$
… … $= 3^{-1}$
… … $=\dfrac{1}{3}$
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Example 4
Simplify $\big(ab^3\big)^2 \times a^{-2}b^{-4} \times \dfrac{1}{a^2b^{-3}}$
Solution:
… … $\big(ab^3\big)^2 \times a^{-2}b^{-4} \times \dfrac{1}{a^2b^{-3}}$
… … $=a^2b^6 \times a^{-2}b^{-4} \times a^{-2}b^3$
… … $=a^{2-2-2} \times b^{6-4+3}$
… … $=a^{-2} \, b^5$
… … $=\dfrac{b^5}{a^2}$
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Example 5
Simplify $\dfrac{5^{3x} \times 25^x}{125^{x+3} \times 5^{-x}}$
Solution:
… … $\dfrac{5^{3x} \times 25^x}{125^{x+3} \times 5^{-x}}$
… … $=\dfrac{5^{3x} \times \big(5^2\big)^x}{\big(5^3\big)^{x+3} \times 5^{-x}}$
… … $= \dfrac{5^{3x} \times\;5^{2x}}{5^{3x+9} \times 5^{-x}}$
… … $= 5^{3x+2x-(3x+9)-(-x)}$
… … $=5^{3x-9}$
Note:
- If the original expression is entered onto the calculator it will not factorise fully due to the different bases.
- By changing everything into base 5 before entering it, we obtain the desired result.
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Example 6
Simplify the following, leaving the answer with positive indices.
… … $\dfrac{1}{x^{-1}+1} - \dfrac{1}{x^{-1}}$
Solution:
… … $\dfrac{1}{x^{-1}+1} - \dfrac{1}{x^{-1}}$
… … $=\Bigg( \dfrac{1}{\frac{1}{x}+1} \times \dfrac{x}{x} \Bigg) - \Big(x\Big)$
- {by multiplying the numerator and denominator of the first fraction by x we avoid fractions within fractions}
… … $=\dfrac{x}{1+x}-x$
- {Now change the second fraction to a common denominator so the two parts can be added}
… … $=\dfrac{x}{1+x}-\dfrac{x(1+x)}{1+x}$
… … $=\dfrac{x-(x+x^2)}{1+x}$
… … $=\dfrac{x-x-x^2}{1+x}$
… … $=\dfrac{-x^2}{1+x}$
Note: On the calculator we have to use the combine function (from the Transformation menu) to get the result as a single fraction.
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