Chapter 10 Exponents and Powers solutions

Exercise 10.1Q1: Evaluate

(i) 3^-2
Sol: (i) To evaluate an expression of the form a^-n,where a is a base and n is a positive exponent, we use the rule 
This means we take the reciprocal of the base raised to the positive exponent. So, for 3^-2 , this would be:

(ii) (-4)^-2
Sol: Applying the same rule to (-4)^-2

Note that the square of −4 is positive because squaring any real number (whether positive or negative) results in a non-negative number.

(iii) (1/2)^-5
Sol: For this expression, we apply the same principle:

In this case, 1/2 raised to the power of 5 gives 1/32 and taking the reciprocal of that gives us 32.
To summarize:

Q2: Simplify and express the result in power notation with positive exponent.
(i) (– 4)⁵ ÷ (– 4)⁸
Sol: When we divide powers with the same base, we subtract the exponents: 

(ii) (1/2³)²
Sol: When we raise a power to another power, we multiply the exponents: 

(iii) 
Sol: When multiplying powers with the same exponent but different bases, we can combine them under the same exponent:

=−(1)×5⁴ =−5⁴
This result cannot be expressed with a positive exponent due to the negative sign in front of 5⁴ .
(iv) 
Sol: First, divide the powers with the same base:

Then, multiply by 3^-5:

To express with a positive exponent:


(v) 2^–3 × (–7)^–3
Sol: Both bases are raised to negative exponents, so we find their reciprocals and multiply: 


Q3: Find the value of 

(i) 
First, evaluate each term inside the parentheses:
3⁰ =1 (Any number raised to the power of 0 is 1.)
(Negative exponents indicate the reciprocal of the base raised to the positive exponent.)
So, the expression becomes:

2² =4, thus:


(ii)
Simplify inside the parentheses first:


For the division:

So, the expression simplifies to:

(iii) 
Evaluate each term:

Summing them up:
4 + 9 + 16 = 29

(iv) 
Any expression raised to the power of 0 is 1, regardless of what the expression inside the parentheses evaluates to. So, without needing to simplify the expression inside, the result is: 1

(v) 
First, evaluate the expression inside the curly braces:

When raising a fraction to a negative exponent, you take the reciprocal of the base and then square it:

Then, raising this result to the power of 2:


Q4: Evaluate. 

(i) 
Sol: First, simplify the numerator and the denominator separately:

So, the expression becomes:

(ii) 
Sol: Simplify each term:

So, the expression becomes:


To summarize:


Q5: Find the value of m for which 

Sol: To find the value of m for which  we can use the properties of exponents to simplify the equation. When dividing numbers with the same base, you subtract the exponents:

So, we have:
5^m+3 = 5⁵
For this equation to be true, the exponents on both sides must be equal (since the bases are the same and non-zero). Therefore, we set the exponents equal to each other:
m + 3 = 5
To solve for m, we subtract 3 from both sides:
m=5−3
m=2
So, the value of m for which the original equation holds true is 2.
Q6: Evaluate. 

Sol: (i)

= (3-4) = -1
(ii)

Q7: Simplify. 

Sol: (i)

(ii) 

Exercise 10.2

Q1. Express the following numbers in standard form.
(i) 0.0000000000085
Sol: This number is 8.5 followed by 11 decimal places to the left, so in standard form, it is:
8.5 × 1 0 ^{− 12}

(ii) 0.00000000000942
Sol: This number is 9.42 followed by 11 decimal places to the left, so in standard form, it is:
9.42 × 1 0 ^{− 12} 

(iii) 6020000000000000
Sol: This number is 6.02 followed by 15 places to the right, so in standard form, it is: 6.02 × 10 ¹⁵

(iv) 0.00000000837
Sol: This number is 8.37 followed by 9 decimal places to the left, so in standard form, it is: 8.37 × 10 ^{− 9}

(v) 31860000000
Sol: This number is 3.186 followed by 10 places to the right, so in standard form, it is:
3.186 × 1 0¹⁰ 

Q2: Express the following numbers in usual form.
(i) 3.02×10⁻⁶
Sol: Shift the decimal point 6 places to the left:
0.00000302

(ii) 4.5 * 104
Sol: Shift the decimal point 4 places to the right:
45,000

(iii) 3 * 10–8
Sol: Shift the decimal point 8 places to the left:
0.00000003

(iv) 1.0001 * 109
Sol: Shift the decimal point 9 places to the right:
1,000,100,000

(v) 5.8 * 1012
Sol: Shift the decimal point 12 places to the right:
5,800,000,000,000

(vi) 3.61492 * 106
Sol: Shift the decimal point 6 places to the right: 
3,614,920

Q3. Express the number appearing in the following statements in standard form.
(i) 1 micron is equal to 
Sol: This can be written as:
1×10⁻⁶m

(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.
Sol: This is:
1.6×10 ⁻¹⁹ coulomb

(iii) Size of a bacteria is 0.0000005 m.
Sol: This can be expressed as:
5×10 ⁻⁷ m

(iv) Size of a plant cell is 0.00001275 m.
Sol: This is:1.275×10 ⁻⁵ m

(v) Thickness of a thick paper is 0.07 mm.
Sol: To express this in meters and in standard form (considering 1 mm = 0.001 m 1 mm=0.001 m):
0.07 mm=0.07×0.001 m=7×10 ⁻² m

Q4: In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?
Sol: To find the total thickness of the stack, we need to sum the thicknesses of all the books and paper sheets.
Books Each book has a thickness of 20 mm. For 5 books, the total thickness is:
5 × 20 mm = 100 mm
Paper Sheets
Each paper sheet has a thickness of 0.016 mm. For 5 paper sheets, the total thickness is:
5×0.016mm=0.08mm
Total Thickness
Adding the thicknesses of the books and paper sheets together gives the total thickness of the stack:
100mm+0.08mm=100.08mm
Therefore, the total thickness of the stack is 100.08 mm.