Exponents of a real number

Exponents of a real number represent the multiplication of the base number. They can be positive, negative, zero or fractional. Some of the types are:

• Positive exponents

• Negative exponents

• Zero exponents

• Rational exponents

We have discussed thoroughly these exponents below.

Positive Exponents

If an exponent indicates how many times a number is multiplied by itself, it is called a Positive exponent.

Example:

3 × 3 × 3 = 33 = 9

Here 33 shows how many times the number 3 is multiplied by itself. So it is a Positive exponent.

Negative Exponents

If an exponent represents the reciprocal of the base raised to the absolute value of the exponent, then it is called a Negative exponent.

Example:

3-3 = $\frac{1}{3^3}=\frac{1}{9}$

Zero Exponent

Any number with an exponent of zero has a value equal to 1.

Example:

50 = 1, 100 = 1, (-3)0 = 0, etc.

Rational Exponents

Rational exponents are also known as Fractional exponents. It represents both roots and powers.

Example: am/n is equivalent to the nth root of a raised to the power m.

163/4 = $(\sqrt[4]{16})^3$ = 23 = 8

Laws of Exponents

Example 1:

Which value among $3^{200},2^{300},$ and $7^{100}$ is the largest?

$3^{200}=(3^{2})^{100}=9^{100}$

$2^{300}=(2^{3})^{100}=8^{100}$

$7^{100}=(7^{1})^{100}=7^{100}$

Thus, the largest number is $3^{200}$.

Example 2:
If $\left (\sqrt{5} \right)^{7}\div \left (\sqrt{5} \right)^{5}=5^{p},$ then the value of $p$ is:

Given: $\frac{(\sqrt{5})^7}{(\sqrt{5})^5} = 5^p$

$⇒(\sqrt{5})^{7 - 5} = 5^p$

$⇒(\sqrt{5})^2 = 5^p$

$⇒5^{\frac{1}{2}×2} = 5^{p}$

$⇒5^1=5^p$

Comparing both sides, we get,

$\therefore p=1$

Laws of Rational Exponents

Rational exponents, also known as fractional exponents, follow similar laws to integer exponents. These laws help simplify and solve expressions involving roots and powers.

Example:

The value of $\frac{(243)^\frac{n}{5}\times 3^{2n+1}}{9^{n}\times 3^{n-1}}$ is:

Given: $\frac{(243)^\frac{n}{5}\times 3^{2n+1}}{9^{n}\times 3^{n-1}}$

= $\frac{(3^5)^\frac{n}{5}\times 3^{2n+1}}{3^{2n}\times 3^{n-1}}$

= $\frac{3^n\times 3^{2n+1}}{3^{2n}\times 3^{n-1}}$

= $\frac{3^{3n+1}}{3^{3n-1}}$

= $\frac{3^{3n}×3^1}{3^{3n}×3^{-1}}$

= ${3^{2}}$

= $9$

Hence, the value is 9.

What is a Surd?

A surd is an irrational number which can not be simplified into a whole number or rational number after the square root or cube root of the number. It is not a perfect square or perfect cube number.

Example: $\sqrt{2}, \sqrt[3]{10}, \sqrt[4]{20},$ etc.

Types of Surds

Simple Surds

Basic irrational numbers that are roots of integers that consist of only one term are called Simple Surds.

Example: $\sqrt2, \sqrt5,\sqrt7,$ etc.

Pure Surds

Surds that are entirely in the root form without any rational number coefficient are called Pure surds. Pure surds can not be written in a fraction form.

Example: $\sqrt{11}, \sqrt{13}, \sqrt{21},$ etc.

Mixed Surds

Surds that are mixed with rational numbers and irrational numbers are called Mixed Surds.

Example: $\sqrt[4]{7}, \sqrt[5]{3}, \sqrt[9]{6},$ etc.

Compound Surds

The sum or difference of two or more surds is called Compound surd.

Example: $(\sqrt3 + \sqrt7 + \sqrt{17}), (\sqrt8 - \sqrt[5]{3}),$ etc

Similar Surds

Surds with the same radicand(number under the roots) are called Similar surds. So it can be easily added or subtracted by combining the exponents.

Example: $2\sqrt5 + 3\sqrt5 = 5\sqrt5$

Binomial Surds

A compound surd consisting of only two terms is called a Binomial Surd.

Example: $(\sqrt6 - \sqrt[5]{9}), (\sqrt5 + \sqrt[2]{12}),$ etc.

Laws of Surds

Simplification Surds

Simplification of surds means converting the surds into a simple form, where the number inside the roots or radicands contains no square factors other than 1.

This process needs some steps which are explained below.

Step 1: Break down the number inside the roots into its prime factors.

Step 2: Pair the prime factors.

Step 3: Take out pairs of prime factors from under the square root, as they become whole numbers.

Example:

Simplify $\sqrt{24}$.

$\sqrt{24}=\sqrt{2×2×2×3}=\sqrt{2^2×2×3}=2\sqrt{2×3}=2\sqrt6$

Simplify $4\sqrt{75}$.

$4\sqrt{75}=4\sqrt{5×5×3}=4\sqrt{5^2×3}=4×5\sqrt{3}=20\sqrt3$

Tips and Tricks

• Any number with an exponent of zero has a value equal to 1.

• am/n is equivalent to the nth root of a raised to the power m.

• $\sqrt a × \sqrt b=\sqrt{ab}$

• Let b be a rational number.

$\frac{1}{\sqrt b}$

Then,

$\frac{1×\sqrt b}{\sqrt b×\sqrt b} = \frac{\sqrt b}{b}$

• Conjugate of $(a+\sqrt b)$ is $(a-\sqrt b)$.

• am × an = amn

• $\frac{a^m}{a^n}$ = am – n

• For surds, simplify the radicand, combine like terms, rationalize denominators, and use surd properties for simplification.

• For Exponents, memorize and apply the laws of exponents, and practice simplifying complex expressions.

Practice Questions

Q1. If $(2^{3})^{2} = 4^{x}$, then $3^{x}$ is equal to:

1. 3

2. 6

3. 9

4. 27

Hint: Use the indices rule: if $a^x = a^y$, then $x=y$

Answer:

Given: $(2^{3})^{2} = 4^{x}$

⇒ $2^{6} = 2^{2x}$

⇒ $6 = 2x$ (If $a^x = a^y$, then $x=y$)

⇒ $x = 3$

$\therefore$ $3^{x} = 3^{3} = 27$

Hence, the correct answer is 27.

Q2. If $4^{(x+y)} = 256$ and $(256)^{(x–y)} = 4$, what are the values of $x$ and $y?$

1. $\frac{17}{8}$ and $\frac{15}{8}$

2. $\frac{17}{4}$ and $\frac{15}{4}$

3. $\frac{9}{17}$ and $\frac{15}{17}$

4. $\frac{8}{17}$ and $\frac{8}{15}$

Hint: Use the indices rule: if $a^x = a^y$, then $x=y$

Answer:

Given: $4^{(x+y)}=256$

⇒ $4^{(x+y)} = 4^4$

⇒ $x+y = 4$

⇒ $x = 4-y$-----(1)

And $(256)^{(x-y)}=4$

⇒ $(4^4)^{(x-y)}=4^1$

⇒ $4x-4y =1$-----(2)

Putting the value of $x$ from equation (1), we get,

$4(4-y)-4y = 1$

⇒ $16-4y-4y =1$

$\therefore y =\frac{15}{8}$

Now, substituting the value of $y$ in equation (1), we get,

$x = 4-\frac{15}{8}$

$\therefore x = \frac{17}{8}$

Hence, the correct answer is $\frac{17}{8}$ and $\frac{15}{8}$.

Q3. The value of $(x^{\frac{1}{3}}+x^{-\frac{1}{3}})(x^{\frac{2}{3}}-1+x^{-\frac{2}{3}})$ is:

1. $x^{-1}+x^{\frac{2}{3}}$

2. $x+x^{-\frac{1}{3}}$

3. $x^{\frac{1}{3}}+x^{–1}$

4. $x+x^{–1}$

Hint: Multiply both the terms and use the rule of indices: $a^x \times a^y = a^{x+y}$

Answer:

Given: $(x^{\frac{1}{3}}+x^{-\frac{1}{3}})(x^{\frac{2}{3}}-1+x^{-\frac{2}{3}})$

$=x^{\frac{1}{3}}×x^{\frac{2}{3}}-x^{\frac{1}{3}}+x^{\frac{1}{3}}×x^{-\frac{2}{3}}+x^{-\frac{1}{3}}×x^{\frac{2}{3}}-x^{-\frac{1}{3}}+x^{-\frac{1}{3}}×x^{-\frac{2}{3}}$

$=x^{\frac{1}{3}+\frac{2}{3}}-x^{\frac{1}{3}}+x^{\frac{1}{3}-\frac{2}{3}}+x^{-\frac{1}{3}+\frac{2}{3}}-x^{-\frac{1}{3}}+x^{-\frac{1}{3}-\frac{2}{3}}$

$=x^1-x^{\frac{1}{3}}+x^{-\frac{1}{3}}+x^{\frac{1}{3}}-x^{-\frac{1}{3}}+x^{-1}$

$=x+x^{-1}$

Hence, the correct answer is $x+x^{-1}$.

Q4. If the numbers $\sqrt[3]{9}, \sqrt[4]{20}$ and $\sqrt[6]{25}$ are arranged in ascending order, then the right arrangement is:

1. $\sqrt[6]{25}<\sqrt[4]{20}< \sqrt[3]{9}$

2. $\sqrt[3]{9}<\sqrt[4]{20}<\sqrt[6]{25}$

3. $\sqrt[4]{20}<\sqrt[6]{25}<\sqrt[3]{9}$

4. $\sqrt[6]{25}<\sqrt[3]{9}<\sqrt[4]{20}$

Hint: Simplify the given values by taking the least common multiple (LCM) of the roots to make roots equal, then compare to find the desired value.

Answer:

Given: The numbers are $\sqrt[3]{9}$, $\sqrt[4]{20}$, $\sqrt[6]{25}$.

LCM of 3, 4, and 6 = 12

$⇒ \sqrt[3]{9}=9^{\frac{1}{3}}=9^{\frac{4}{12}}=\sqrt[12]{6561}$

$⇒ \sqrt[4]{20}=20^{\frac{1}{4}}=20^{\frac{3}{12}}=\sqrt[12]{8000}$

$⇒ \sqrt[6]{25}=25^{\frac{1}{6}}=25^{\frac{2}{12}}=\sqrt[12]{625}$

By comparing these values, we get:

$\sqrt[12]{625}<\sqrt[12]{6561}<\sqrt[12]{8000}$

$⇒\sqrt[6]{25}<\sqrt[3]{9}<\sqrt[4]{20}$

Hence, the correct answer is $\sqrt[6]{25}<\sqrt[3]{9}<\sqrt[4]{20}$.

Q5. What is the value of $\left[\frac{12}{(\sqrt5+\sqrt3)}+\frac{18}{(\sqrt{5}-\sqrt3)}\right]$?

1. $15(\sqrt5–\sqrt3)$

2. $3(5\sqrt5+\sqrt3)$

3. $15(\sqrt5+\sqrt3)$

4. $3(3\sqrt5+\sqrt3)$

Hint: Use the formula: $(\sqrt a+\sqrt b)(\sqrt a–\sqrt b) = a-b$

Answer:

Given:

$\left[\frac{12}{(\sqrt5+\sqrt3)}+\frac{18}{(\sqrt{5}-\sqrt3)}\right]$

= $\frac{12\sqrt5-12\sqrt 3+18\sqrt5+18\sqrt 3}{(\sqrt5+\sqrt3)(\sqrt{5}-\sqrt3)}$

= $\frac{30\sqrt5+6\sqrt3}{5-3}$

= $\frac{6(5\sqrt5+\sqrt3)}{2}$

= $3(5\sqrt5+\sqrt3)$

Hence, the correct answer is $3(5\sqrt5+\sqrt3)$.

Q6. The quotient when $10^{100}$ is divided by $5^{75}$ is:

1. $2^{25}×10^{75}$

2. $10^{25}$

3. $2^{75}$

4. $2^{75}×10^{25}$

Hint: Solve by using the formula: $\frac{a^{m}}{a^{n}}=a^{(m-n)}$ and simplify to find the unknown value.

Answer:

Given: $10^{100}$ is divided by $5^{75}$.

= $\frac{10^{100}}{5^{75}}$

= $\frac{(2×5)^{100}}{5^{75}}$

= $\frac{2^{100}×5^{100}}{5^{75}}$

= $2^{100}×\frac{5^{100}}{5^{75}}$

By using, $\frac{a^{m}}{a^{n}}=a^{(m–n)}$

= $2^{100}×5^{(100-75)}$

= $2^{100}×5^{25}$

= $2^{75}×2^{25}×5^{25}$

= $2^{75}×10^{25}$

Hence, the correct answer is $2^{75}×10^{25}$.

Q7. What is the value of the given expression?

$\frac{4^{a + 4}- 5 \times 4^{a + 2}}{15 \times 4^a - 2^2 \times 4^a}$

1. 16

2. 64

3. 20

4. 24

Hint: Use the information below to solve the problem.

$x^{ab}=x^a×x^b$

Answer:

$\frac{4^{a+ 2}(4^{2} - 5 \times 1)}{4^{a}(15 - 4)}$

$=\frac{4^{a + 2}(11)}{4^{a}(11)}$

$=\frac{4^{a}\times4^{2}(11)}{4^{a}(11)}$

$=4^2$

$=16$

Hence, the correct answer is 16.

Q8. If $a=b^p,b=c^q,c=a^r$, then $pqr$ is:

1. 1

2. 0

3. -1

4. abc

Hint: Use the values of $b$ and $c$ in the equation $a=b^p$.

Answer:

Given:

$a=b^p,b=c^q,c=a^r$

$a=b^p$

⇒ $a=(c^q)^p$

⇒ $a=(a^r)^{pq}$

⇒ $a^1=(a)^{pqr}$

$\therefore pqr=1$

Hence, the correct answer is 1.

Q9. When xm is multiplied by xn product is 1. The relation between m and n is:

1. mn = 1

2. m = n

3. m + n = 1

4. m = – n

Hint: Use the following formula to solve the question:

xm × xn = xm + n

Answer:

xm × xn = 1

⇒ xm + n = x0

⇒ m + n = 0

⇒ m = –n

Hence, the correct answer is m = – n.

Q10. If $2^{2x-y}=16$ and $2^{x+y}=32$, the value of $xy$ is:

1. 2

2. 4

3. 6

4. 8

Hint: If am = an, then m = n.

Answer:

Given: $2^{2x-y}=16$

$⇒2^{2x-y}=2^4$

$⇒2x-y=4$--------------------(1)

Again, $2^{x+y}=32$

$⇒2^{x+y}=2^5$

$⇒x+y=5$ ---------------------(2)

Solving equations (1) and (2), we get:

$x=3$ and $y= 2$

$\therefore xy=3×2=6$

Hence, the correct answer is $6$.