03Aindices

toc = Index Laws =

Recall, the following laws for simplifying indices.

math \\ . \qquad a^m \times a^n = a^{m+n} \\. \\ . \qquad \dfrac{a^m}{a^n} = a^{m - n} \\. \\ . \qquad (a^m)^n = a^{mn} math . math \\ . \qquad a^0 = 1 \\. \\ . \qquad a^{-m} = \dfrac{1}{a^m} \\. \\ . \qquad a^ {\frac{1}{n}} = \sqrt[n]{a} \\. \\ math . math . \qquad a^{\frac{m}{n}} =\sqrt[n]{a^m} = \big( \sqrt[n]{a} \big)^m math

For a revision of these laws, go here:
 * basic index laws
 * negative indices
 * fractional indices

**Example 1**
math . \qquad \text{Simplify } \; 15(x^2y^{-2})^4 \div \left( 3(x^4y)^{-2} \right) math

__Solution:__

math . \qquad 15 \big( x^2y^{-2} \big)^4 \div \left( 3 \big( x^4y \big)^{-2} \right) math . math . \qquad = \dfrac{15x^8 y^{-8}} {3x^{-8} y^{-2}} \\. \\ . \qquad =5x^{16}y^{-6} \\. \\ .\qquad =\dfrac{5x^{16}}{y^6} math

**Example 2**
math . \qquad \text{Find } \; 125^{\frac{2}{3}} math

__Solution:__

math . \qquad 125^{\frac{2}{3}} math . math \\ . \qquad = \big( \sqrt[3]{125} \big)^2 \\. \\ . \qquad =5^2 \\. \\ . \qquad =25 math

__alternate solution__ math \\ . \qquad 125^{\frac{2}{3}} \\. \\ . \qquad =\big(5^3\big)^\frac{2}{3} math . math \\ . \qquad =5^{(3\times\frac{2}{3})} \\. \\ . \qquad =5^2 \\. \\ . \qquad =25 math

**Example 3**
math . \qquad \text{Find }\; 81^{-0.25} math

__Solution:__

math \\ . \qquad 81^{-0.25} \\. \\ . \qquad =\dfrac{1}{81^{0.25}} \\. \\ . \qquad =\dfrac{1}{81^\frac{1}{4}} math . math \\ . \qquad =\dfrac{1}{\big(3^4\big)^\frac{1}{4}} \\. \\ . \qquad =\dfrac{1}{3} math

**Example 4**
math . \qquad \text{Simplify } \; \big(ab^3\big)^2 \times a^{-2}b^{-4} \times \dfrac{1}{a^2b^{-3}} math

__Solution:__ math \\ . \qquad \big(ab^3\big)^2 \times a^{-2}b^{-4} \times \dfrac{1}{a^2b^{-3}} \\. \\ . \qquad =a^2b^6 \times a^{-2}b^{-4} \times a^{-2}b^3 \\. \\ . \qquad =a^{2-2-2} \times b^{6-4+3} \\. \\ . \qquad =a^{-2} \, b^5 \\. \\ . \qquad =\dfrac{b^5}{a^2} math

**Example 5**
math . \qquad \text{Simplify } \dfrac{5^{3x} \times 25^x}{125^{x+3} \times 5^{-x}} math

__Solution__:

math \\ . \qquad \dfrac{5^{3x} \times 25^x}{125^{x+3} \times 5^{-x}} \\. \\ . \qquad =\dfrac{5^{3x} \times \big(5^2\big)^x}{\big(5^3\big)^{x+3} \times 5^{-x}} \\. \\ . \qquad = \dfrac{5^{3x} \times\;5^{2x}}{5^{3x+9} \times 5^{-x}} \\. \\ . \qquad = 5^{3x+2x-(3x+9)-(-x)} \\. \\ . \qquad =5^{3x-9} math

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.

**Example 6**
... Simplify the following, leaving the answer with positive indices. math . \qquad \quad \dfrac{1}{x^{-1}+1} - \dfrac{1}{x^{-1}} math

__Solution: __

math \\ . \qquad \dfrac{1}{x^{-1}+1} - \dfrac{1}{x^{-1}} \\. \\ . \qquad =\Bigg( \dfrac{1}{\frac{1}{x}+1} \times \dfrac{x}{x} \Bigg) - \Big(x\Big) math

{by multiplying the numerator and denomenator of the first fraction by x we avoid fractions within fractions} math . \qquad =\dfrac{x}{1+x}-x math

{Now change the second fraction to a common denominator so the two parts can be added} math \\ . \qquad =\dfrac{x}{1+x}-\dfrac{x(1+x)}{1+x} \\. \\ . \qquad =\dfrac{x-(x+x^2)}{1+x} \\. \\ . \qquad =\dfrac{x-x-x^2}{1+x} \\. \\ . \qquad =\dfrac{-x^2}{1+x} math

Note: On the calculator we have to use the **combine** function (from the **Transformation** menu) to get the result as a single fraction.

Go to top of page flat

.