04 4indexlaws

# 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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