Exponents - Exercise 5(B)

3² + 2³ is equal to :

1. 15
2. 16
3. 17
4. 18

Answer

3² = 3 x 3 = 9

2³ = 2 x 2 x 2 = 8.

3² + 2³ = 9 + 8 = 17.

Hence, option 3 is the correct option.

(7 - 5)⁵ is equal to :

1. 6
2. 8
3. 16
4. 32

Answer

We have:

(7 - 5)⁵

⇒ (7 - 5) = 2

2⁵ = 2 x 2 x 2 x 2 x 2 = 32

Hence, option 4 is the correct option.

(-5)⁴ is equal to :

1. 20
2. -125
3. 625
4. -1024

Answer

Concept:

(-5)⁴ = (-5) x (-5) x (-5) x (-5) = 625.

Hence, option 3 is the correct option.

$\Big(\dfrac{-3}{5}\Big)^5$ is equal to :

1. $\dfrac{-243}{3125}$

2. $\dfrac{-81}{625}$

3. $\dfrac{81}{625}$

4. $\dfrac{512}{1875}$

Answer

If the base is negative and the exponent is odd, the result is negative.

$\Big(\dfrac{-3}{5}\Big)^5 \\[1em] = \dfrac{(-3)^5}{5^5} \\[1em] = \dfrac{(-3) \times (-3) \times (-3) \times (-3) \times (-3)}{5 \times 5 \times 5 \times 5 \times 5} \\[1em] = \dfrac{-243}{3125}$

Hence, option 1 is the correct option.

The value of x such that $\Big(\dfrac{3}{7}\Big)^3 \times \Big(\dfrac{3}{7}\Big)^{-8} = \Big(\dfrac{3}{7}\Big)^{2x + 3}$ is

1. -4
2. -2
3. 0
4. 1

Answer

Given:

$\Big(\dfrac{3}{7}\Big)^3 \times \Big(\dfrac{3}{7}\Big)^{-8} = \Big(\dfrac{3}{7}\Big)^{2x + 3}$

LHS = $\Big(\dfrac{3}{7}\Big)^3 \times \Big(\dfrac{3}{7}\Big)^{-8}$

$= \dfrac{3^3}{7^3} \times \dfrac{3^{-8}}{7^{-8}} \\[1em] = \dfrac{3^3 \times 3^{-8}}{7^3 \times 7^{-8}} \\[1em] = \dfrac{3^{3+(-8)}}{7^{3+(-8)}} = \dfrac{3^{3-8}}{7^{3-8}} \\[1em] = \dfrac{3^{-5}}{7^{-5}} \\[1em] = \Big(\dfrac{3}{7}\Big)^{-5} \\[1em]$

Now, LHS = $\Big(\dfrac{3}{7}\Big)^{-5}$

As the base of both LHS and RHS is same, let us compare the exponents:

-5 = 2x + 3
2x = -5 - 3
2x = -(5 + 3)
2x = -8
x = $\dfrac{-8}{2}$
x = -4

Hence, option 1 is the correct option.

The value of $\Big[\Big\lbrace\Big(\dfrac{-1}{3}\Big)^{-2}\Big\rbrace^{-1}\Big]^{2}$ is

1. 81

2. -81

3. $\dfrac{1}{81}$

4. $-\dfrac{1}{81}$

Answer

Given:

$\Big[\Big\lbrace\Big(\dfrac{-1}{3}\Big)^{-2}\Big\rbrace^{-1}\Big]^{2}$

Multiply the exponents: (-2) x (-1) x 2 = 4. $\quad$ [Power of power rule]

∴ $\Big(\dfrac{-1}{3}\Big)^4 = \dfrac{(-1)^4}{3^4} = \dfrac{1}{81}$

Hence, option 3 is the correct option.

The number 34613000 when expressed in exponential form is equal to

1. 3461.3 x 10³
2. 3.4613 x 10⁷
3. 0.34613 x 10⁹
4. 34.613 x 10⁵

Answer

Move the decimal 7 places to the left: 3.4613 x 10⁷.

Verification: 3.4613 x 10,000,000 = 34613000.

Hence, option 2 is the correct option.

Which is larger ?

(i) 2³ or 3²

(ii) 2⁵ or 5²

(iii) 3⁷ or 7³.

Answer

(i) 2³ or 3²

2³ = 2 x 2 x 2 = 8

3² = 3 x 3 = 9

Clearly, 8 < 9.

Hence, 3² is larger.

(ii) 2⁵ or 5²

2⁵ = 2 x 2 x 2 x 2 x 2 = 32

5² = 5 x 5 = 25

Clearly, 32 > 25.

Hence, 2⁵ is larger.

(iii) 3⁷ or 7³.

3⁷ = 3 x 3 x 3 x 3 x 3 x 3 x 3 = 2187

7³ = 7 x 7 x 7 = 343

Clearly, 2187 > 343.

Hence, 3⁷ is larger.

Fill in the blanks :

(i) In an exponential form x^m; x is called the ............... .

(ii) We have : $\dfrac{a^m}{a^n} = \dfrac{1}{a^{n-m}}$ if ............... .

(iii) Any number to the power 0 is equal to ............... .

(iv) The exponential form is also called ............... .

(v) The reciprocal of x^a is equal to x to the power ............... .

Answer

(i) In an exponential form x^m; x is called the base.

(ii) We have : $\dfrac{a^m}{a^n} = \dfrac{1}{a^{n-m}}$ if n > m.

(iii) Any number to the power 0 is equal to 1.

(iv) The exponential form is also called power notation.

(v) The reciprocal of x^a is equal to x to the power -a.

State True or False :

(i) For any non-zero number a, we have (a^m)ⁿ = a^m+n.

(ii) If a and b are non-zero numbers, then $\Big\lbrace\Big(\dfrac{a}{b}\Big)^m\Big\rbrace^n = \Big(\dfrac{b}{a}\Big)^{-mn}$

(iii) For a non-zero number x; (x^m x xⁿ) is equal to x to the power (m + n).

(iv) If a number p is multiplied n times, then the resulting number is pⁿ.

(v) In an exponential notation xⁿ; n is called the index.

Answer

(i) False
Reason — The power of a power law states that (a^m)ⁿ = a^{m x n}. The expression a^m+n is the result of multiplying powers with the same base (a^m x aⁿ).

(ii) True
Reason — According to the power of a power rule, $\Big\lbrace\left(\dfrac{a}{b}\right)^m\Big\rbrace^n = \left(\dfrac{a}{b}\right)^{mn}$. By applying the reciprocal rule $\left(\dfrac{a}{b}\right)^{mn} = \left(\dfrac{b}{a}\right)^{-mn}$, the statement is mathematically correct.

(iii) True
Reason — This is the product law of exponents, which states that for any non-zero base x, x^m x xⁿ = x^(m+n).

(iv) True
Reason — By definition, exponential notation is a shorthand for repeated multiplication; if p is multiplied n times, it is written as pⁿ.

(v) True
Reason — In the notation xⁿ, the number n is commonly referred to as the exponent, power, or index.

A computer purchased for ₹72900 loses two-third of its value every year. Its value is evaluated at the end of every year.

(1) Which of the following expressions gives the value of the computer (in ₹) after n years?

1. $\dfrac{72900}{\Big(\dfrac{2}{3}\Big)^{n}}$

2. $\dfrac{2^{n} \times 27900}{3^{n}}$

3. $\dfrac{72900}{3^{n}}$

4. $\dfrac{72900}{2^{n} \times 3^{n}}$

(2) Find the value of the computer after 3 years.

1. ₹8100
2. ₹2430
3. ₹5600
4. ₹2700

(3) In how many years will the value of the computer be less than ₹200 ?

1. 6 years
2. 7 years
3. 8 years
4. 10 years

(4) By how much will the value of the computer reduce in 4 years ?

1. ₹24300
2. ₹1800
3. ₹8100
4. ₹72000

Answer

(1) Given:

Initial Value = ₹72,900

Loss in value every year = $\dfrac{2}{3}$

Value remaining every year = $1 - \dfrac{2}{3} = \dfrac{1}{3}$ of the previous year's value.

Every year, the value is multiplied by $\dfrac{1}{3}$. After n years, the value is $72900 \times (\dfrac{1}{3})^n$, which is $\dfrac{72900}{3^n}$.

Hence, option 3 is the correct option.

(2) Value of the computer after 3 years = ?

Value of the computer after n years = $\dfrac{72900}{3^n} \quad$ [From previous step]

By replacing the value of 'n' with 3, we get:

₹ $\dfrac{72900}{3^3} = ₹ \dfrac{72900}{27}$

= ₹2700

Hence, option 4 is the correct option.

(3) We know at 3 years, value = ₹2700. $\quad$ [From previous step]

Value remaining every year = $1 - \dfrac{2}{3} = \dfrac{1}{3}$ of the previous year's value. $\quad$ [From step 1]

∴ Value after 4 years = $\dfrac{1}{3} \times 2700 = \dfrac{2700}{3} = ₹ 900$

Value after 5 years = $\dfrac{1}{3} \times 900 = \dfrac{900}{3} = ₹ 300$

Value after 6 years = $\dfrac{1}{3} \times 300 = \dfrac{300}{3} = ₹ 100$

Since 100 < 200, it takes 6 years.

Hence, option 1 is the correct option.

(4) Value of the computer reduced in 4 years = ?

Value after 4 years = ₹ $\dfrac{72900}{3^4} = ₹ \dfrac{72900}{81} = ₹ 900$

Value of the computer reduced in 4 years = Initial value - Value after 4 years

Substituting the values in above, we get:

Value of the computer reduced in 4 years = ₹72900 - ₹ 900 = ₹ 72000

Hence, option 4 is the correct option.

In a bacteria culture under observation in a laboratory, the population of 50 bacteria doubles itself every hour.

(1) Which of the following expressions gives the bacterial population after n hours ?

1. $\dfrac{50}{2^{n}}$

2. 25ⁿ

3. 50 x 2ⁿ

4. $\dfrac{50^{n}}{2}$

(2) The population size of the bacteria after 3 hours will be

1. 200
2. 300
3. 400
4. 500

(3) How many bacteria will be there in the culture after 1 day ?

1. $\dfrac{50}{2^{12}}$

2. 50 x 2²⁴

3. 50 x 2¹²

4. $\dfrac{50}{2^{24}}$

(4) If the culture is observed after every one hour, find the number of hours after which the population size of the bacteria will be larger than 1000.

1. 5
2. 6
3. 8
4. 10

Answer

(1) Given:

Initial Population = 50

Growth Rate = Doubles every hour (x2)

The population starts at 50 and multiplies by 2 for every hour.

General formula after n hours:

Population = 50 × 2ⁿ

Hence, option 3 is the correct option.

(2) Population size after 3 hours = ?

Population size after n hours = 50 × 2ⁿ $\quad$ [From step 1]

By replacing the value of 'n' with 3, we get:

Population size after 3 hours = 50 × 2³ = 50 x 8 = 400

Hence, option 3 is the correct option.

(3) Bacteria in the culture after 1 day = ?

We know that 1 day has 24 hours

Population size after n hours = 50 × 2ⁿ $\quad$ [From step 1]

By replacing the value of 'n' with 24, we get:

Bacteria in the culture after 1 day = 50 × 2²⁴

Hence, option 2 is the correct option.

(4) After how many hours will the population be larger than 1000?

Let's test hours (n):

n = 4: 50 x 2⁴ = 50 x 16 = 800

n = 5: 50 x 2⁵ = 50 x 32 = 1600

Since 1600 > 1000, it happens at 5 hours.

Hence, option 1 is the correct option.

Assertion: $\dfrac{x^{4}}{y^{3}}$ can also be written as $\dfrac{x+x+x+x}{y+y+y}$.

Reason: In x^m, x is called the base and m is called the exponent.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
3. Assertion (A) is true but Reason (R) is false.
4. Assertion (A) is false but Reason (R) is true.

Answer

Assertion (A) is false but Reason (R) is true.

Explanation

Assertion is false because, the expression $\dfrac{x^4}{y^3}$ represents repeated multiplication, not addition. Specifically, $x^4 = x \times x \times x \times x$ and $y^3 = y \times y \times y$. Adding the variables as shown in the assertion is incorrect.

Reason is true as it is the correct mathematical definition for the components of an exponential term.

Hence, option 4 is the correct option.

Assertion: x^m ÷ y^m = (x ÷ y)^m

Reason: a^m x b^m = (ab)^m

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
3. Assertion (A) is true but Reason (R) is false.
4. Assertion (A) is false but Reason (R) is true.

Answer

Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Explanation

Assertion is true because, it is the Power of a Quotient rule. It states that when two different bases are divided and raised to the same power, the power can be applied to the quotient.

Reason is true because, it is the Power of a Product rule. It correctly states that a^m x b^m = (ab)^m.

While both are valid laws of exponents, the rule for multiplication (Reason) does not explain the rule for division (Assertion).

Hence, option 2 is the correct option.

Assertion: 2⁰ + 3⁰ + 4⁰ = (2 + 3 + 4)⁰

Reason: x⁰ = 1.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
3. Assertion (A) is true but Reason (R) is false.
4. Assertion (A) is false but Reason (R) is true.

Answer

Assertion (A) is false but Reason (R) is true.

Explanation

Assertion is false.

Let's evaluate both sides using the zero exponent rule:

LHS: 2⁰ + 3⁰ + 4⁰ = 1 + 1 + 1 = 3.

RHS: (2 + 3 + 4)⁰ = (9)⁰ = 1

Since $3 \neq 1$, the assertion is false.

Reason is true because, the rule x⁰ = 1 (for any non-zero x) is a fundamental law of exponents.

Hence, option 4 is the correct option.

2³ + 2³ + 2³ + 2³ is equal to:

1. 2⁵
2. 2¹²
3. 281
4. 216

Answer

Given expression: 2³ + 2³ + 2³ + 2³

Since 2³ is being added 4 times, we can write it as:

= 4 × 2³

= 2² × 2³ $\hspace{2cm}$[∵ 4 = 2²]

= 2^{2 + 3} $\hspace{2cm}$[∵ a^m × aⁿ = a^{m + n}]

= 2⁵

∴ 2³ + 2³ + 2³ + 2³ = 2⁵.

Hence, option 1 is the correct option.

The value of $\dfrac{\left(-\dfrac{1}{2}\right)^5}{\left(-\dfrac{1}{2}\right)^4} \div \dfrac{\left(-\dfrac{1}{8}\right)}{\left(-\dfrac{1}{4}\right)}$ is:

1. 2
2. 0
3. 1
4. −1

Answer

Step 1: Simplify the first fraction

$\dfrac{\left(-\dfrac{1}{2}\right)^5}{\left(-\dfrac{1}{2}\right)^4} \\[1em] = \left(-\dfrac{1}{2}\right)^{5 - 4} \quad [∵ \dfrac{a^m}{a^n} = a^{m - n}, \text{ if } m \gt n] \\[1em] = \left(-\dfrac{1}{2}\right)^1 \\[1em] = -\dfrac{1}{2}$

Step 2: Simplify the second fraction

$\dfrac{\left(-\dfrac{1}{8}\right)}{\left(-\dfrac{1}{4}\right)} = \left(-\dfrac{1}{8}\right) \times \left(-\dfrac{4}{1}\right) \quad \text{[Reciprocal of } -\dfrac{1}{4} \text{ is } -\dfrac{4}{1}] \\[1em] = \dfrac{4}{8} \\[1em] = \dfrac{1}{2}$

Step 3: Divide the two results

$-\dfrac{1}{2} \div \dfrac{1}{2} = -\dfrac{1}{2} \times \dfrac{2}{1} \quad \text{[Reciprocal of } \dfrac{1}{2} \text{ is } \dfrac{2}{1}] \\[1em] = -\dfrac{2}{2} \\[1em] = -1$

Hence, option 4 is the correct option.

The value of $\left(\dfrac{a^p}{a^q}\right) \times \left(\dfrac{a^q}{a^r}\right) \times \left(\dfrac{a^r}{a^p}\right)$ is:

1. 1

2. $a^{\dfrac{1}{pqr}}$

3. a^pqr

4. a

Answer

Given expression: $\left(\dfrac{a^p}{a^q}\right) \times \left(\dfrac{a^q}{a^r}\right) \times \left(\dfrac{a^r}{a^p}\right)$

Rearranging the factors,

= $\dfrac{a^p \times a^q \times a^r}{a^q \times a^r \times a^p}$

= 1

for a ≠ 0.

∴ The value of the given expression is 1.

Hence, option 1 is the correct option.

The value of a^b − b^a, if a = 3 and b = 7 is:

1. 1825
2. 1840
3. 1844
4. 1850

Answer

Given:

a = 3 and b = 7

Substituting the values of a and b in the given expression, we get:

a^b − b^a = 3⁷ − 7³

Step 1: Evaluate 3⁷

3⁷ = 3 × 3 × 3 × 3 × 3 × 3 × 3 = 2187

Step 2: Evaluate 7³

7³ = 7 × 7 × 7 = 343

Step 3: Subtract

3⁷ − 7³ = 2187 − 343 = 1844

∴ The value of a^b − b^a is 1844.

Hence, option 3 is the correct option.

The value of $\dfrac{3^4}{2^3} + \dfrac{\left(\dfrac{1}{9} \times 4\right)^2}{\left(\dfrac{-1}{3}\right)^3 \div \left(\dfrac{1}{-2}\right)^3}$ is:

1. $\dfrac{81}{24}$

2. $\dfrac{259}{24}$

3. $\dfrac{81}{8}$

4. $\dfrac{259}{8}$

Answer

Step 1: Evaluate the first term

$\dfrac{3^4}{2^3} = \dfrac{3 \times 3 \times 3 \times 3}{2 \times 2 \times 2} = \dfrac{81}{8}$

Step 2: Evaluate the numerator of the second term

$\left(\dfrac{1}{9} \times 4\right)^2 = \left(\dfrac{4}{9}\right)^2 = \dfrac{4^2}{9^2} = \dfrac{16}{81}$

Step 3: Evaluate the denominator of the second term

$\left(\dfrac{-1}{3}\right)^3 = \dfrac{(-1)^3}{3^3} = \dfrac{-1}{27} \\[1em] \left(\dfrac{1}{-2}\right)^3 = \dfrac{1^3}{(-2)^3} = \dfrac{-1}{8} \\[1em] \left(\dfrac{-1}{3}\right)^3 \div \left(\dfrac{1}{-2}\right)^3 = \dfrac{-1}{27} \div \dfrac{-1}{8} \\[1em] = \dfrac{-1}{27} \times \dfrac{8}{-1} \quad \text{[Reciprocal of } \dfrac{-1}{8} \text{ is } \dfrac{8}{-1}] \\[1em] = \dfrac{8}{27}$

Step 4: Simplify the second term

$\dfrac{\dfrac{16}{81}}{\dfrac{8}{27}} = \left(\dfrac{16}{81} \times \dfrac{27}{8}\right) = \left(\dfrac{2}{3} \times \dfrac{1}{1}\right) = \dfrac{2}{3}$

Step 5: Add the two terms

$\dfrac{81}{8} + \dfrac{2}{3} = \dfrac{81 \times 3 + 2 \times 8}{24} \quad \text{[LCM of 8 and 3 is 24]} \\[1em] = \dfrac{243 + 16}{24} \\[1em] = \dfrac{259}{24}$

∴ The value of the given expression is $\dfrac{259}{24}$.

Hence, option 2 is the correct option.

The value of (−8) × (−8) × (−8) × (−8) × (−8) × (−8) × (−8) × (−8) × (−8) × (−8) is:

1. −2(8)⁵
2. −(2)¹⁶
3. −(4)⁸
4. none of these

Answer

The given expression has (−8) multiplied by itself 10 times.

So, (−8) × (−8) × (−8) × (−8) × (−8) × (−8) × (−8) × (−8) × (−8) × (−8) = (−8)¹⁰

Since 10 is an even number, the result is positive.
[∵ A negative number raised to an even power is positive]

(−8)¹⁰ = 8¹⁰ = (2³)¹⁰ = 2³⁰ $\hspace{2cm}$[∵ (a^m)ⁿ = a^mn]

Now, let us check each option:

Option 1: −2(8)⁵ = −2 × 2¹⁵ = −2¹⁶ (negative value)

Option 2: −(2)¹⁶ (negative value)

Option 3: −(4)⁸ = −(2²)⁸ = −2¹⁶ (negative value)

Since 2³⁰ is positive and all the given options are negative, none of them is equal to (−8)¹⁰.

∴ The value of the given expression does not match any of options 1, 2 or 3.

Hence, option 4 is the correct option.

The value of (p^x × p^y) ÷ (p^x ÷ p^y) is:

1. 2p^x
2. p^2x
3. 2p^y
4. p^2y

Answer

Given expression: (p^x × p^y) ÷ (p^x ÷ p^y)

Rewrite the division in the second bracket as a fraction:

= $(p^x \times p^y) \div \left(\dfrac{p^x}{p^y}\right)$

Dividing by a fraction means multiplying by its reciprocal:

= $p^x \times p^y \times \dfrac{p^y}{p^x}$

Cancel p^x from the numerator and denominator:

= p^y × p^x

= p^{y + y} $\quad$ [using a^m × aⁿ = a^{m + n}]

= p^2y

∴ The value of the given expression is p^2y.

Hence, option 4 is the correct option.

The value of x^y − y^x + xy, if x = 6, y = 2 is:

1. −12
2. 12
3. −16
4. 16

Answer

Given:

x = 6 and y = 2

Substituting the values of x and y in the given expression, we get:

x^y − y^x + xy = 6² − 2⁶ + (6 × 2)

Step 1: Evaluate 6²

6² = 6 × 6 = 36

Step 2: Evaluate 2⁶

2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64

Step 3: Evaluate 6 × 2

6 × 2 = 12

Step 4: Substitute and simplify

6² − 2⁶ + 6 × 2 = 36 − 64 + 12

= 48 − 64

= −16

∴ The value of x^y − y^x + xy is −16.

Hence, option 3 is the correct option.

If 3^x = 500, then the value of 3^x−2 is:

1. $\dfrac{500}{9}$

2. $\dfrac{1000}{9}$

3. $\dfrac{500}{3}$

4. $\dfrac{100}{9}$

Answer

Given:

3^x = 500

We need to find the value of 3^{x − 2}.

Using the quotient law of exponents:

3^{x − 2} = $\dfrac{3^x}{3^2}$ $\hspace{2cm}$[∵ a^{m − n} = $\dfrac{a^m}{a^n}$]

Substituting the value of 3^x, we get:

3^{x − 2} = $\dfrac{500}{3^2}$

= $\dfrac{500}{9}$ $\hspace{2cm}$[∵ 3² = 9]

∴ The value of 3^{x − 2} is $\dfrac{500}{9}$.

Hence, option 1 is the correct option.

Rajat claims, "A negative number raised to a power is always less than the number itself." Rajat's statement is:

1. true
2. false
3. can't say anything
4. none of these

Answer

Rajat claims, “A negative number raised to a power is always less than the number itself.”

To check the statement, consider a negative number raised to an even power.

For example,

(−2)² = (−2) × (−2) = 4

But,

4 > −2

So, a negative number raised to a power is not always less than the number itself.

∴ Rajat's statement is false.

Hence, option 2 is the correct option.

{(9²)⁴ × 9⁵} ÷ 9⁸ is equal to:

1. 9
2. 9⁸
3. 9⁵
4. 9⁹

Answer

Given expression: {(9²)⁴ × 9⁵} ÷ 9⁸

Step 1: Simplify (9²)⁴

(9²)⁴ = 9^{2 × 4} = 9⁸ $\hspace{2cm}$[∵ (a^m)ⁿ = a^mn]

Step 2: Simplify the bracket

9⁸ × 9⁵ = 9^{8 + 5} = 9¹³ $\hspace{2cm}$[∵ a^m × aⁿ = a^{m + n}]

Step 3: Divide by 9⁸

9¹³ ÷ 9⁸ = 9^{13 − 8} = 9⁵ $\hspace{2cm}$[∵ a^m ÷ aⁿ = a^{m − n}]

∴ The value of the given expression is 9⁵.

Hence, option 3 is the correct option.