Exponents and Powers

FAMILY - Every person has 2 biological parents. Study the family tree above.

1.0Introduction

• How many 2 s are multiplied to determine the number of great grandparents?
• How many 2 s would you multiply to determine the number of great-great grandparents?

An expression like $2\times 2\times 2\times 2$ can be written as the power $2^4$.

The table below shows how to write and read powers.

Powers with negative exponents

Definition : In general for non-zero integer $a,a^{-n}=\frac{1}{a^n}$ where n is a positive integer. $a^{-n}$ is the multiplicative inverse of ${\mathrm{a}}^{\mathrm{n}}$.

• Q. Find the value of (i) $4^{-3}$ (ii) $\frac{1}{7^{-2}}$ (iii) ${\left(\frac{-6}{7}\right)}^{-3}$ Solution: (i) $4^{-3}=\frac{1}{4^3}=\frac{1}{64}$ (ii) $\frac{1}{7^{-2}}=7^2=49$ (iii) ${\left(\frac{-6}{7}\right)}^{-3}=\frac{1}{{\left(\frac{-6}{7}\right)}^3}=\frac{1}{\frac{(-6{)}^3}{7^3}}=\frac{7^3}{(-6{)}^3}=\frac{343}{-216}=-\frac{343}{216}$
• Q. Write the reciprocal of (i) $4^3$ (ii) ${\left(\frac{6}{5}\right)}^2$ (iii) ${\left(\frac{-3}{2}\right)}^4$ Solution: (i) Reciprocal of $4^3=\frac{1}{4^3}$ or $4^{-3}$ (ii) Reciprocal of ${\left(\frac{6}{5}\right)}^2=\frac{1}{{\left(\frac{6}{5}\right)}^2}=\frac{1}{\frac{6^2}{5^2}}=\frac{5^2}{6^2}={\left(\frac{5}{6}\right)}^2$ or ${\left(\frac{6}{5}\right)}^{-2}$ (iii) Reciprocal of ${\left(\frac{-3}{2}\right)}^4=\frac{1}{{\left(\frac{-3}{2}\right)}^4}=\frac{1}{\frac{(-3{)}^4}{(2{)}^4}}=\frac{(2{)}^4}{(-3{)}^4}={\left(\frac{2}{-3}\right)}^4$ or ${\left(\frac{-3}{2}\right)}^{-4}$
• Q. Simplify : $\left[{\left(\frac{1}{4}\right)}^{-2}-{\left(\frac{1}{2}\right)}^{-3}\right]÷{\left(\frac{1}{4}\right)}^{-2}$ Solution:

• Q. Rewrite the following using positive exponents. Assume that no denominators are equal to 0 . (i) $b^{-6}$ (ii) ${\mathbf{c}}^{-9}{\mathrm{d}}^3$ (iii) $7{\mathbf{x}}^{-4}{\mathbf{y}}^2$ Solution: (i) ${\mathrm{b}}^{-6}=\frac{1}{{\mathrm{b}}^6}$ (ii) ${\mathrm{c}}^{-9}{\mathrm{d}}^3=\frac{1}{{\mathrm{c}}^9}\cdot {\mathrm{d}}^3=\frac{{\mathrm{d}}^3}{{\mathrm{c}}^9}$ (iii) $7x^{-4}{y}^2=7\cdot \frac{1}{x^4}\cdot y^2=\frac{7y^2}{x^4}$

2.0Laws of exponents

You have studied the laws of exponents, with exponents as positive integer. These laws also hold true for negative exponents.

Let us see if these hold true for integral exponents i.e. exponents which can be negative also. (i) ${\mathrm{X}}^{\mathrm{m}}\times {\mathrm{X}}^{\mathrm{n}}={\mathrm{X}}^{\mathrm{m}+\mathrm{n}}$ Let us check the first law by an example $2^{-2}\times 2^{-3}=\frac{1}{2^2}\times \frac{1}{2^3}=\frac{1}{2^2\times 2^3}=\frac{1}{2^{3+2}}=\frac{1}{2^5}=2^{-5}$ or $2^{(-2)+(-3)}=2^{-5}$ So, first law holds for integral exponents

(ii) ${\mathrm{x}}^{\mathrm{m}}÷{\mathrm{x}}^{\mathrm{n}}={\mathrm{x}}^{\mathrm{m}-\mathrm{n}}$ Let us check the second law by an example $4^{-1}÷4^{-2}=\frac{1}{4}÷\frac{1}{4^2}=\frac{1}{4}÷\frac{1}{4\times 4}=\frac{1}{4}\times \frac{4\times 4}{1}=\frac{4}{1}=4=4^1$ or $4[(-1)-(-2)]=4(-1+2)=4^1$ So, the second law holds for integral exponents

(iii) ${\left({\mathrm{x}}^{\mathrm{m}}\right)}^{\mathrm{n}}={\mathrm{x}}^{mn}$ Let us check this law by an example ${\left(8^{-1}\right)}^{-3}={\left(\frac{1}{8^1}\right)}^{-3}={\left(\frac{8}{1}\right)}^3=8^3$ or $8(-1)\times (-3)=8^3$ So, the third law holds for integral exponents.

(iv) ${\mathrm{x}}^{\mathrm{m}}\times {\mathrm{y}}^{\mathrm{m}}=(\mathrm{xy}{)}^{\mathrm{m}}$ Let us check this by an example $2^{-2}\times 3^{-2}=\frac{1}{2^2}\times \frac{1}{3^2}=\frac{1}{2^2\times 3^2}=\frac{1}{(2\times 3{)}^2}=(2\times 3{)}^{-2}$ or $2^{-2}\times 3^{-2}=(2\times 3{)}^{-2}$ So, the fourth law holds for integral exponents.

(v) $\frac{x^m}{y^m}={\left(\frac{x}{y}\right)}^m$ Let us check this by an example $$\begin{gathered} \frac{3^{-2}}{5^{-2}}=\frac{5^2}{3^2}={\left(\frac{5}{3}\right)}^2={\left(\frac{3}{5}\right)}^{-2} \\ \text{ or }\frac{3^{-2}}{5^{-2}}={\left(\frac{3}{5}\right)}^{-2} \end{gathered}$$ So, the fifth law holds for integral exponents

• $2^0\ne 0;2^0=1$
• ${\mathrm{X}}^{\mathrm{m}}+{\mathrm{X}}^{\mathrm{n}}\ne {\mathrm{X}}^{\mathrm{m}+\mathrm{n}};{\mathrm{X}}^{\mathrm{m}}\times {\mathrm{X}}^{\mathrm{n}}={\mathrm{X}}^{\mathrm{m}+\mathrm{n}}$ $x^m-x^n\ne x^{m-n};x^m÷x^n=x^{m-n}$
• Q. Simplify and write the answer in an exponential form. (i) ${\left(3^7÷3^{11}\right)}^3\times 3^{-8}$ (ii) ${\left[{\left(\frac{6}{7}\right)}^3\right]}^2\times {\left[{\left(\frac{6}{7}\right)}^{-4}\right]}^2$ Solution: (i) ${\left(3^7÷3^{11}\right)}^3\times 3^{-8}$

(ii) ${\left[{\left(\frac{6}{7}\right)}^3\right]}^2\times {\left[{\left(\frac{6}{7}\right)}^{-4}\right]}^2={\left(\frac{6}{7}\right)}^{3\times 2}\times {\left(\frac{6}{7}\right)}^{-4\times 2}$ $={\left(\frac{6}{7}\right)}^6\times {\left(\frac{6}{7}\right)}^{-8}={\left(\frac{6}{7}\right)}^{6+(-8)}={\left(\frac{6}{7}\right)}^{-2}={\left(\frac{7}{6}\right)}^2$

• Q. Find the value of $x$. ${\left(\frac{-11}{7}\right)}^{-5}\times {\left(\frac{-11}{7}\right)}^x={\left(\frac{-11}{7}\right)}^3$ Solution: $$\begin{gathered} {\left(\frac{-11}{7}\right)}^{-5}\times {\left(\frac{-11}{7}\right)}^x={\left(\frac{-11}{7}\right)}^3 \\ \Rightarrow {\left(\frac{-11}{7}\right)}^{(-5)+x}={\left(\frac{-11}{7}\right)}^3 \end{gathered}$$ Since, the base is same on both sides of the expression, their exponents should also be the same. $$\begin{gathered} \Rightarrow -5+x=3\Rightarrow x=3+5 \\ \therefore x=8 \end{gathered}$$
• Q. Find the value of $\frac{x}{y}$, if ${\left(\frac{4}{9}\right)}^{-10}\times {\left(\frac{18}{7}\right)}^{-10}={\left(\frac{x}{y}\right)}^{-10}$ Solution: ${\left(\frac{4}{9}\right)}^{-10}\times {\left(\frac{18}{7}\right)}^{-10}={\left(\frac{x}{y}\right)}^{-10}$ $\Rightarrow {\left(\frac{4}{9}\times \frac{18}{7}\right)}^{-10}={\left(\frac{x}{y}\right)}^{-10}\Rightarrow {\left(\frac{8}{7}\right)}^{-10}={\left(\frac{x}{y}\right)}^{-10}$ Hence, $\frac{x}{y}=\frac{8}{7}$

ASTRONOMY: The diameter of the Sun is approximately $20^9$ metres. A model of the Sun has a diameter of ${20}^{-1}$ meters. How many models would fit across the diameter of the Sun? Explanation: $20^9÷20^{-1}=20^{9-(-1)}=20^{10}$ models

Writing expanded form of a decimal number using exponential notation

Look at the expanded form of 96829.653 $96829.653=9\times 10000+6\times 1000+8\times 100+2\times 10+9\times 1+6\times \frac{1}{10}+5\times \frac{1}{100}+3\times \frac{1}{1000}$ We can express this expansion in exponential notation using exponents of 10 . Therefore, $96829.653=9\times 10^4+6\times 10^3+8\times 10^2+2\times 10^1+9\times 10^0+6\times 10^{-1}+5\times 10^{-2}+3\times 10^{-3}$ We observe that, the exponents of 10 start from the highest value and go on decreasing by 1 at each step, from left to right.

• Q. Expand the following number using exponents: 1256.249 Solution: $1256.249=1\times 1000+2\times 100+5\times 10+6\times 1+\frac{2}{10}+\frac{4}{100}+\frac{9}{1000}$ $=1\times 10^3+2\times 10^2+5\times 10^1+6\times 10^0+2\times 10^{-1}+4\times 10^{-2}+9\times 10^{-3}$
• Q. Form the number from the given expanded form : $7\times 10^3+4\times 10^1+2\times 10^0+6\times 10^{-2}+8\times 10^{-3}$ Explanation: $7\times 10^3+4\times 10^1+2\times 10^0+\frac{6}{10^2}+\frac{8}{10^3}$ $=7\times 1000+4\times 10+2\times 1+\frac{6}{100}+\frac{8}{1000}$ $=7000+40+2+0.06+0.008$ $=7042.068$

3.0Scientific notation

The Sun is located at a distance of $1,000,000,000,000,000,000\mathrm{km}$ from the center of milky way. The average size of an atom is about 0.00000003 centimetre across. The length of these numbers makes them awkward to work with. Scientific notation is a shorthand way of writing such numbers. To express any number in scientific notation, write it as the product of a power of ten and a number greater than or equal to 1 but less than 10 . In scientific notation, the Sun is located at a distance of $1.0\times 10^{21}\mathrm{m}$ from the centre of our galaxy and the size of each atom is $3.0\times 10^{-8}$ centimetres across.

• Q. Write the numbers in the standard form (i) 0.00925 (ii) 457000000 (iii) 0.32458 Solution: (i) $0.00925=9.25\times 10^{-3}$ (ii) $457000000=4.57\times 10^8$ (iii) $0.32458=3.2458\times 10^{-1}$

4.0Comparing very large and very small numbers

To compare two numbers written in scientific notation, first compare the powers of ten. The number with the greater power of ten is greater. If the powers of ten are the same, compare the values between one and ten. $2.7\times 10^{13}>2.7\times 10^9$ $\left(\because 10^{13}>10^9\right)$ $3.98\times 10^{22}>2.52\times 10^{22}$

• Q. The thickness of a sheet of paper is $1.6\times 10^{-3}\mathrm{cm}$ and the thickness of a human hair is $5\times 10^{-3}\mathbf{cm}$. Compare the two. Solution: $\frac{\text{ Thickness of hair }}{\text{ Thickness of paper }}=\frac{5\times 10^{-3}}{1.6\times 10^{-3}}=\frac{5\times 10^{-3}\times 10^3}{1.6}=\frac{5}{1.6}=3.125$ The hair is approximately three times thicker than the paper.
• Q. ASTRONOMY The distance between the Sun and the Moon is $1.49984\times 10^{11}\mathrm{m}$ and the distance between the Moon and the Earth is $3.84\times 10^8\mathrm{m}$. During Lunar eclipse, the Earth comes between the Sun and the Moon. Find out the distance between the Sun and the Earth at this time.
  
  Explanation: Distance between The Sun and the Moon $=1.49984\times 10^{11}\mathrm{m}$ The Moon and the Earth $=3.84\times 10^8\mathrm{m}$ The Sun and the Earth $=\left(1.49984\times 10^{11}-3.84\times 10^8\right)\mathrm{m}$ $=\left(1.49984\times 1000\times 10^8-3.84\times 10^8\right)\mathrm{m}$ $=\left(1499.84\times 10^8-3.84\times 10^8\right)\mathrm{m}$ $=(1499.84-3.84)\times 10^8\mathrm{m}$ $=1496\times 10^8\mathrm{m}$

5.0Mind map