# 8th Class Mathematics Exponents and Power Laws of Exponent

## Laws of Exponent

Category : 8th Class

### Laws of Exponent

There are various laws of exponents. They are laws of addition, laws of multiplication and laws of division.

(i) ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$

(ii) $\frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$

(iii) ${{a}^{m}}\times {{b}^{m}}={{(a\times b)}^{m}}$

(iv) ${{\left[ {{\left( \frac{a}{b} \right)}^{n}} \right]}^{m}}={{\left( \frac{a}{b} \right)}^{mn}}$

(v) ${{\left( \frac{a}{b} \right)}^{-n}}={{\left( \frac{b}{a} \right)}^{n}}$

(vi) ${{\left( \frac{a}{b} \right)}^{0}}=1$

(vii) ${{(ab)}^{n}}={{a}^{n}}{{b}^{n}}$

Important Points to keep in Mind

•  ${{x}^{0}}=1$, where $x\ne 0$. $x$ can be anything (except zero), including numbers, variables, or an equation.
•  ${{x}^{1}}=x$
•  ${{x}^{-n}}=\frac{1}{{{x}^{n}}}$, Where $x\ne 0$

(i) ${{2}^{3}}{{2}^{5}}={{2}^{3+5}}={{2}^{8}}$

(ii) ${{w}^{2}}{{w}^{3}}={{w}^{5}}$

(iii) $x{{y}^{2}}{{x}^{3}}{{y}^{3}}{{x}^{4}}{{y}^{4}}={{x}^{8}}{{y}^{9}}$

While working with exponents there are certain rules that we need to remember.

${{\text{4}}^{\text{2}}}\times {{\text{4}}^{\text{5}}}=4\text{7}$

It means:$\text{4}\times \text{4}\times \text{4}\times \text{4}\times \text{4}\times \text{4}\times \text{4}$ or $\text{4}\text{.4}\text{.4}\text{.4}\text{.4}\text{.4}\text{.4}$

Add the exponent, if base are same.

Uses of Exponents

The exponents can be used for various purposes such as comparing large and small numbers, expressing large and small numbers in the standard forms. It is used to express the distance between any two celestial bodies which cannot be expressed in the form of normal denotion. It is also useful in writing the numbers in scientific notation. The size of the microorganisms is very-very small and it cannot be written in normal denotion and can easily be expressed in exponential form.

Radicals are the fractional exponents of any number. Index of the radical becomes the denominator of the fractional power.

$\sqrt[n]{a}=\frac{1}{{{a}^{n}}}$

i.e.  $\sqrt{9}=\sqrt[2]{9}={{9}^{\frac{1}{2}}}=3$

Express $\sqrt[\mathbf{3}]{\mathbf{2}}\,\sqrt[\mathbf{4}]{\mathbf{2}}$ as a Single Radical Term

Let us convert the radicals to exponential expressions, and then apply laws of exponent to combine the factors:

$\sqrt[3]{2}\,\,\sqrt[4]{2}={{2}^{\frac{1}{3}}}\,\,{{2}^{\frac{1}{4}}}={{2}^{\frac{1}{3}+\frac{1}{4}}}={{2}^{\frac{7}{12}}}=\sqrt[12]{{{2}^{7}}}$

Simplify $\frac{\sqrt{5}}{\sqrt[3]{5}}$

Solution:

$\frac{{{5}^{\frac{1}{2}}}}{{{5}^{\frac{1}{3}}}}={{5}^{\frac{1}{2}.\frac{1}{3}}}={{5}^{\frac{1}{6}}}$

${{\left( \frac{2}{3} \right)}^{4}}=\left( \frac{2}{3} \right)\times \left( \frac{2}{3} \right)\times \left( \frac{2}{3} \right)\times \left( \frac{2}{3} \right)$

Solution:

$=\frac{2\times 2\times 2\times 2}{3\times 3\times 3\times 3}=\frac{{{2}^{4}}}{{{3}^{4}}}=\frac{16}{81}$

Expand: ${{\left( \frac{x}{10} \right)}^{5}}$

Solution: Raising the top and bottom numbers to the power of 5 gives:

${{\left( \frac{x}{10} \right)}^{5}}=\frac{{{x}^{5}}}{{{10}^{5}}}=\frac{{{x}^{5}}}{100000}$

•  Zero raised to the power is not always zero.
•  The positive quantity raised to the power of negative index is always positive.
•  Zero was discovered by the Babylonians in Mesopotamia in around 300 B.C.
•  The rule of mathematics given by Brahma gupta is known as Brahmas phutasiddhanta.
•  The number which appears maximum number of times in the decimal expansion of pie upto the 6 billion decimal places.

${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$

${{({{a}^{m}})}^{n}}={{a}^{mn}}$

${{(ab)}^{n}}={{a}^{n}}{{b}^{n}}$

${{\left( \frac{a}{b} \right)}^{n}}=\frac{{{a}^{n}}}{{{b}^{n}}}$

${{a}^{0}}=1$

${{a}^{-n}}=\frac{1}{{{a}^{n}}}$

Simplify: ${{(-5)}^{3}}$

(a) -125

(b) -120

(c) -105

(d) -121

(e) None of these

Explanation:

$(-5)(-5)(-5)=-125$

Number -5 has exponent 3. Therefore, option (a) is correct and rest of the options is incorrect.

Simplify: $-{{(3)}^{3}}-{{(-3)}^{2}}+{{(-2)}^{2}}$

(a) -31

(b) -21

(c) -32

(d) -33

(e) None of these

Explanation:

$-(3\times 3\times 3)-(-3\times -3)+(-2\times -2)=-27-9+4=-32$

Find the value of x such that ${{\left( \frac{64}{125} \right)}^{2}}{{\left( \frac{4}{5} \right)}^{4}}{{\left( \frac{16}{25} \right)}^{2x+1}}={{\left( \frac{256}{625} \right)}^{3x}}$

(a) $\frac{3}{2}$

(b) $\frac{2}{3}$

(c) $\frac{1}{3}$

(d) $\frac{1}{2}$

(e) None of these

By what number ${{\left( -\frac{4}{3} \right)}^{-5}}$ must be multiplied so that the result is $\frac{16}{9}$.

(a) ${{\left( \frac{4}{3} \right)}^{5}}$

(b) ${{\left( \frac{4}{3} \right)}^{7}}$

(c) ${{\left( \frac{4}{3} \right)}^{-5}}$

(d) ${{\left( \frac{4}{3} \right)}^{-7}}$

(e) None of these

The scientific notation of 165000000000000 is given by:

(a) $16.5\times {{10}^{13}}$

(b) $165\times {{10}^{12}}$

(c) $1650\times {{10}^{11}}$

(d) $1.65\times {{10}^{14}}$

(e) None of these