Selina Concise Solutions for ICSE Class 7 Mathematics Chapter 5 Exponents Including Laws of Exponents

Exercise 5(A)

Question 1. Find the value of:
(i) \( 6^2 \)
(ii) \( 7^3 \)
(iii) \( 4^4 \)
(iv) \( 5^5 \)
(v) \( 8^3 \)
(vi) \( 7^5 \)
Answer:
(i) \( 6^2 = 6 \times 6 = 36 \)
(ii) \( 7^3 = 7 \times 7 \times 7 = 343 \)
(iii) \( 4^4 = 4 \times 4 \times 4 \times 4 = 256 \)
(iv) \( 5^5 = 5 \times 5 \times 5 \times 5 \times 5 = 3125 \)
(v) \( 8^3 = 8 \times 8 \times 8 = 512 \)
(vi) \( 7^5 = 7 \times 7 \times 7 \times 7 \times 7 = 16807 \)
In simple words: To find the value, multiply the bottom number by itself as many times as the top power tells you to do.

Exam Tip: Avoid the common mistake of multiplying the base number directly by its exponent. For example, remember that \( 6^2 \) is \( 6 \times 6 \), not \( 6 \times 2 \).

Question 2. Evaluate:
(i) \( 2^3 \times 4^2 \)
(ii) \( 2^3 \times 5^2 \)
(iii) \( 3^3 \times 5^2 \)
(iv) \( 2^2 \times 3^3 \)
(v) \( 3^2 \times 5^3 \)
(vi) \( 5^3 \times 2^4 \)
(vii) \( 3^2 \times 4^2 \)
(viii) \( (4 \times 3)^3 \)
(ix) \( (5 \times 4)^2 \)
Answer:
(i) \( 2^3 \times 4^2 = 2 \times 2 \times 2 \times 4 \times 4 = 8 \times 16 = 128 \)
(ii) \( 2^3 \times 5^2 = 2 \times 2 \times 2 \times 5 \times 5 = 8 \times 25 = 200 \)
(iii) \( 3^3 \times 5^2 = 3 \times 3 \times 3 \times 5 \times 5 = 27 \times 25 = 675 \)
(iv) \( 2^2 \times 3^3 = 2 \times 2 \times 3 \times 3 \times 3 = 4 \times 27 = 108 \)
(v) \( 3^2 \times 5^3 = 3 \times 3 \times 5 \times 5 \times 5 = 9 \times 125 = 1125 \)
(vi) \( 5^3 \times 2^4 = 5 \times 5 \times 5 \times 2 \times 2 \times 2 \times 2 = 125 \times 16 = 2000 \)
(vii) \( 3^2 \times 4^2 = 3 \times 3 \times 4 \times 4 = 9 \times 16 = 144 \)
(viii) \( (4 \times 3)^3 = (4 \times 3) \times (4 \times 3) \times (4 \times 3) = 12 \times 12 \times 12 = 1728 \)
(ix) \( (5 \times 4)^2 = (5 \times 4) \times (5 \times 4) = 20 \times 20 = 400 \)
In simple words: Write out each of the exponent terms as repeated multiplication, calculate their individual values, and then multiply them together.

Exam Tip: For expressions inside brackets like \( (4 \times 3)^3 \), it is usually easier to multiply the numbers inside the brackets first and then find the power of that final number.

Question 3. Evaluate:
(i) \( \left(\frac{3}{4}\right)^4 \)
(ii) \( \left(-\frac{5}{6}\right)^5 \)
(iii) \( \left(\frac{-3}{-5}\right)^3 \)
Answer:
(i) \( \left(\frac{3}{4}\right)^4 = \left(\frac{3}{4}\right) \times \left(\frac{3}{4}\right) \times \left(\frac{3}{4}\right) \times \left(\frac{3}{4}\right) = \frac{3 \times 3 \times 3 \times 3}{4 \times 4 \times 4 \times 4} = \frac{81}{256} \)
(ii) \( \left(-\frac{5}{6}\right)^5 = \left(-\frac{5}{6}\right) \times \left(-\frac{5}{6}\right) \times \left(-\frac{5}{6}\right) \times \left(-\frac{5}{6}\right) \times \left(-\frac{5}{6}\right) = \frac{(-5) \times (-5) \times (-5) \times (-5) \times (-5)}{6 \times 6 \times 6 \times 6 \times 6} = -\frac{3125}{7776} \)
(iii) \( \left(\frac{-3}{-5}\right)^3 = \left(\frac{-3}{-5}\right) \times \left(\frac{-3}{-5}\right) \times \left(\frac{-3}{-5}\right) = \frac{(-3) \times (-3) \times (-3)}{(-5) \times (-5) \times (-5)} = \frac{-27}{-125} = \frac{27}{125} \)
In simple words: To find the power of a fraction, multiply the top number by itself and the bottom number by itself as many times as the power says.

Exam Tip: Remember that an odd power of a negative number will always yield a negative result, whereas an even power of a negative number becomes positive.

Question 4. Evaluate:
(i) \( \left(\frac{2}{3}\right)^3 \times \left(\frac{3}{4}\right)^2 \)
(ii) \( \left(-\frac{3}{4}\right)^3 \times \left(\frac{2}{3}\right)^4 \)
(iii) \( \left(\frac{3}{5}\right)^2 \times \left(-\frac{2}{3}\right)^3 \)
Answer:
(i) \( \left(\frac{2}{3}\right)^3 \times \left(\frac{3}{4}\right)^2 = \left(\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}\right) \times \left(\frac{3}{4} \times \frac{3}{4}\right) = \frac{8}{27} \times \frac{9}{16} = \frac{1}{6} \)
(ii) \( \left(-\frac{3}{4}\right)^3 \times \left(\frac{2}{3}\right)^4 = \left(-\frac{3}{4} \times -\frac{3}{4} \times -\frac{3}{4}\right) \times \left(\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}\right) = \frac{-27}{64} \times \frac{16}{81} = -\frac{1}{2} \)
(iii) \( \left(\frac{3}{5}\right)^2 \times \left(-\frac{2}{3}\right)^3 = \left(\frac{3}{5} \times \frac{3}{5}\right) \times \left(-\frac{2}{3} \times -\frac{2}{3} \times -\frac{2}{3}\right) = \frac{9}{25} \times \left(\frac{-8}{27}\right) = -\frac{8}{75} \)
In simple words: Expand each fractional power, multiply the numerators and denominators, and then simplify your final fraction by cancelling common factors.

Exam Tip: It is usually helpful to cancel common terms in the numerator and denominator before multiplying them fully, as this keeps the calculations much smaller and easier to manage.

Question 5. Which is greater:
(i) \( 2^3 \) or \( 3^2 \)
(ii) \( 2^5 \) or \( 5^2 \)
(iii) \( 4^3 \) or \( 3^4 \)
(iv) \( 5^4 \) or \( 4^5 \)
Answer:
To compare these values, we will evaluate each exponent separately:
(i) For the first pair:
\( 2^3 = 2 \times 2 \times 2 = 8 \)
\( 3^2 = 3 \times 3 = 9 \)
Since 9 is larger than 8, we find that \( 3^2 > 2^3 \).
(ii) For the second pair:
\( 2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32 \)
\( 5^2 = 5 \times 5 = 25 \)
Since 32 is larger than 25, we find that \( 2^5 > 5^2 \).
(iii) For the third pair:
\( 4^3 = 4 \times 4 \times 4 = 64 \)
\( 3^4 = 3 \times 3 \times 3 \times 3 = 81 \)
Since 81 is larger than 64, we find that \( 3^4 > 4^3 \).
(iv) For the fourth pair:
\( 5^4 = 5 \times 5 \times 5 \times 5 = 625 \)
\( 4^5 = 4 \times 4 \times 4 \times 4 \times 4 = 1024 \)
Since 1024 is larger than 625, we find that \( 4^5 > 5^4 \).
In simple words: To see which number is larger, work out the full value of each power and then compare the two final numbers.

Exam Tip: Do not just look at the bases and exponents to guess which is larger; write down the full calculations to show your step-by-step reasoning clearly.

Question 6. Express each of the following in exponential form:
(i) 512
(ii) 1250
(iii) 1458
(iv) 3600
(v) 1350
(vi) 1176
Answer:
(i) Prime factorization of 512:

\( 512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^9 \)

(ii) Prime factorization of 1250:

\( 1250 = 2 \times 5 \times 5 \times 5 \times 5 = 2 \times 5^4 \)

(iii) Prime factorization of 1458:

\( 1458 = 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 2 \times 3^6 \)

(iv) Prime factorization of 3600:

\( 3600 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 = 2^4 \times 3^2 \times 5^2 \)

(v) Prime factorization of 1350:

\( 1350 = 2 \times 3 \times 3 \times 3 \times 5 \times 5 = 2 \times 3^3 \times 5^2 \)

(vi) Prime factorization of 1176:

\( 1176 = 2 \times 2 \times 2 \times 3 \times 7 \times 7 = 2^3 \times 3 \times 7^2 \)
In simple words: Break each number down by dividing it by prime numbers until you get to 1, and then count up how many of each prime number you used to write the final exponent.
Exam Tip: Always make sure to divide using prime numbers, starting with the smallest factor (such as 2 or 3) and moving to larger ones systematically.

Question 7. If \( a = 2 \) and \( b = 3 \), find the value of:
(i) \( (a + b)^2 \)
(ii) \( (b - a)^3 \)
(iii) \( (a \times b)^a \)
(iv) \( (a \times b)^b \)
Answer:
Substitute the given values \( a = 2 \) and \( b = 3 \) into each expression:
(i) \( (a + b)^2 = (2 + 3)^2 = 5^2 = 5 \times 5 = 25 \)
(ii) \( (b - a)^3 = (3 - 2)^3 = 1^3 = 1 \times 1 \times 1 = 1 \)
(iii) \( (a \times b)^a = (2 \times 3)^2 = 6^2 = 6 \times 6 = 36 \)
(iv) \( (a \times b)^b = (2 \times 3)^3 = 6^3 = 6 \times 6 \times 6 = 216 \)
In simple words: Replace the letters with their given numbers, work out the calculation inside the brackets first, and then apply the exponent.

Exam Tip: Double check that you substitute the correct values for \( a \) and \( b \) into both the base and the exponent, and complete the work inside parentheses first.

Question 8. Express:
(i) 1024 as a power of 2.
(ii) 343 as a power of 7.
(iii) 729 as a power of 3.
Answer:
(i) Repeatedly dividing 1024 by 2:

\( 1024 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{10} \)

(ii) Repeatedly dividing 343 by 7:

\( 343 = 7 \times 7 \times 7 = 7^3 \)

(iii) Repeatedly dividing 729 by 3:

\( 729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6 \)
In simple words: Divide the big number by the requested base number until you reach 1, and count how many steps it took to get your final power.
Exam Tip: Draw clear division ladders to avoid errors when counting the number of repeated divisions.

Question 9. If \( 27 \times 32 = 3^x \times 2^y \); find the values of \( x \) and \( y \).
Answer:
First, find the prime factors of 27:

\( 27 = 3 \times 3 \times 3 = 3^3 \).
Comparing \( 27 = 3^x \implies 3^3 = 3^x \implies x = 3 \).

Next, find the prime factors of 32:

\( 32 = 2 \times 2 \times 2 \times 2 \times 2 = 2^5 \).
Comparing \( 32 = 2^y \implies 2^5 = 2^y \implies y = 5 \).

By comparing both sides of the equation, we get \( x = 3 \) and \( y = 5 \).
In simple words: Write both numbers in exponent form. Match the power of 3 to find x, and match the power of 2 to find y.
Exam Tip: Equate only the exponents of the same base on both sides of the equation to find the values of \( x \) and \( y \) correctly.

Question 10. If \( 64 \times 625 = 2^a \times 5^b \); find:
(i) the values of \( a \) and \( b \).
(ii) \( 2^b \times 5^a \)
Answer:
(i) Find the prime factors of 64:

\( 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6 \).
So, \( 64 = 2^a \implies 2^6 = 2^a \implies a = 6 \).

Next, find the prime factors of 625:

\( 625 = 5 \times 5 \times 5 \times 5 = 5^4 \).
So, \( 625 = 5^b \implies 5^4 = 5^b \implies b = 4 \).
Therefore, \( a = 6 \) and \( b = 4 \).

(ii) Now, we substitute \( a = 6 \) and \( b = 4 \) into the expression \( 2^b \times 5^a \):
\( 2^4 \times 5^6 = (2 \times 2 \times 2 \times 2) \times (5 \times 5 \times 5 \times 5 \times 5 \times 5) \)
\( = 16 \times 15625 \)
\( = 250000 \)
In simple words: Find the powers of the base numbers first to get the values of a and b. Then swap the powers in the second part and multiply the values to find your final answer.
Exam Tip: Be careful not to mix up the letters; note that the second part of the question asks for \( 2^b \times 5^a \), meaning you must use the exponent \( b \) on base 2 and exponent \( a \) on base 5.

Exercise 5(B)

Question 1. Fill in the blanks:
(i) In \( 5^2 = 25 \), base = ......... and index = ..........
(ii) If index = \( 3x \) and base = \( 2y \), the number = .........
Answer:
(i) In \( 5^2 = 25 \), base = \( 5 \) and index = \( 2 \)
(ii) If index = \( 3x \) and base = \( 2y \), the number = \( (2y)^{3x} \)
In simple words: The base is the big number at the bottom, and the index is the small number on top. When writing a number with a base and an index, put the base in brackets to keep it together.

Exam Tip: Always put the base in brackets when it has both a number and a letter, like \( 2y \), before writing the exponent.

Question 2. Evaluate:
(i) \( 2^8 \div 2^3 \)
(ii) \( 2^3 \div 2^8 \)
(iii) \( (2^6)^0 \)
(iv) \( (3^0)^6 \)
(v) \( 8^3 \times 8^{-5} \times 8^4 \)
(vi) \( 5^4 \times 5^3 \div 5^5 \)
(vii) \( 5^4 \div 5^3 \times 5^5 \)
(viii) \( 4^4 \div 4^3 \times 4^0 \)
(ix) \( (3^5 \times 4^7 \times 5^8)^0 \)
Answer:
(i) \( 2^8 \div 2^3 = \frac{2^8}{2^3} = 2^{8-3} = 2^5 = 32 \)
(ii) \( 2^3 \div 2^8 = \frac{2^3}{2^8} = 2^{3-8} = 2^{-5} = \frac{1}{2^5} = \frac{1}{32} \)
(iii) \( (2^6)^0 = 2^{6 \times 0} = 2^0 = 1 \)
(iv) \( (3^0)^6 = 3^{0 \times 6} = 3^0 = 1 \)
(v) \( 8^3 \times 8^{-5} \times 8^4 = 8^{3 + (-5) + 4} = 8^{7-5} = 8^2 = 64 \)
(vi) \( 5^4 \times 5^3 \div 5^5 = \frac{5^4 \times 5^3}{5^5} = 5^{4+3-5} = 5^2 = 25 \)
(vii) \( 5^4 \div 5^3 \times 5^5 = 5^{4-3} \times 5^5 = 5^1 \times 5^5 = 5^{1+5} = 5^6 = 15625 \)
(viii) \( 4^4 \div 4^3 \times 4^0 = 4^{4-3} \times 1 = 4^1 \times 1 = 4 \)
(ix) \( (3^5 \times 4^7 \times 5^8)^0 = 1 \)
In simple words: When you divide powers with the same base, subtract the exponents. When you multiply them, add the exponents. Any non-zero number with an exponent of 0 is just 1.

Exam Tip: Remember that any base raised to the power of 0 is always 1, no matter how big or complex the terms inside the brackets are.

Question 3. Simplify, giving Solutions with positive index:
(i) \( 2b^6 \cdot b^3 \cdot 5b^4 \)
(ii) \( x^2 y^3 \cdot 6x^5 y \cdot 9x^3 y^4 \)
(iii) \( (-a^5)(a^2) \)
(iv) \( (-y)^2 (-y)^3 \)
(v) \( (-3)^2 (3)^3 \)
(vi) \( (-4x)(-5x^2) \)
(vii) \( (5a^2 b)(2ab^2)(a^3 b) \)
(viii) \( x^{2a+7} \cdot x^{2a-8} \)
(ix) \( 3^y \cdot 3^2 \cdot 3^{-4} \)
(x) \( 2^{4a} \cdot 2^{3a} \cdot 2^{-a} \)
(xi) \( 4x^2 y^2 \div 9x^3 y^3 \)
(xii) \( (10^2)^3 (x^8)^{12} \)
(xiii) \( (a^{10})^{10} (1^6)^{10} \)
(xiv) \( (n^2)^2 (-n^2)^3 \)
(xv) \( -(3ab)^2 (-5a^2 b c^4)^2 \)
(xvi) \( (-2)^2 \times (0)^3 \times (3)^3 \)
(xvii) \( (2a^3)^4 (4a^2)^2 \)
(xviii) \( (4x^2 y^3)^3 \div (3x^2 y^3)^3 \)
(xix) \( \left(\frac{1}{2x}\right)^3 \times (6x)^2 \)
(xx) \( \left(\frac{1}{4ab^2c}\right)^2 \div \left(\frac{3}{2a^2bc^2}\right)^4 \)
(xxi) \( \frac{(5x^7)^3 \cdot (10x^2)^2}{(2x^6)^7} \)
(xxii) \( \frac{(7p^2 q^9 r^5)^2 (4pqr)^3}{(14p^6 q^{10} r^4)^2} \)
Answer:
(i) \( 2b^6 \cdot b^3 \cdot 5b^4 = (2 \times 5) \cdot b^{6+3+4} = 10b^{13} \)
(ii) \( x^2 y^3 \cdot 6x^5 y \cdot 9x^3 y^4 = (6 \times 9) \cdot x^{2+5+3} \cdot y^{3+1+4} = 54x^{10}y^8 \)
(iii) \( (-a^5)(a^2) = - (a^{5+2}) = -a^7 \)
(iv) \( (-y)^2 (-y)^3 = (-y)^{2+3} = (-y)^5 = -y^5 \)
(v) \( (-3)^2 (3)^3 = 9 \times 27 = 243 = 3^5 \)
(vi) \( (-4x)(-5x^2) = (-4 \times -5) \cdot x^{1+2} = 20x^3 \)
(vii) \( (5a^2 b)(2ab^2)(a^3 b) = (5 \times 2) \cdot a^{2+1+3} \cdot b^{1+2+1} = 10a^6 b^4 \)
(viii) \( x^{2a+7} \cdot x^{2a-8} = x^{2a+7+2a-8} = x^{4a-1} \)
(ix) \( 3^y \cdot 3^2 \cdot 3^{-4} = 3^{y+2-4} = 3^{y-2} \)
(x) \( 2^{4a} \cdot 2^{3a} \cdot 2^{-a} = 2^{4a+3a-a} = 2^{6a} \)
(xi) \( 4x^2 y^2 \div 9x^3 y^3 = \frac{4x^2 y^2}{9x^3 y^3} = \frac{4}{9} x^{2-3} y^{2-3} = \frac{4}{9} x^{-1} y^{-1} = \frac{4}{9xy} \)
(xii) \( (10^2)^3 (x^8)^{12} = 10^{2 \times 3} \cdot x^{8 \times 12} = 10^6 x^{96} \)
(xiii) \( (a^{10})^{10} (1^6)^{10} = a^{100} \cdot 1^{60} = a^{100} \cdot 1 = a^{100} \)
(xiv) \( (n^2)^2 (-n^2)^3 = n^4 \times (-1)^3 (n^2)^3 = n^4 \times (-n^6) = -n^{10} \)
(xv) \( -(3ab)^2 (-5a^2 b c^4)^2 = -(9a^2 b^2) \times (25 a^4 b^2 c^8) = -225a^6 b^4 c^8 \)
(xvi) \( (-2)^2 \times (0)^3 \times (3)^3 = 4 \times 0 \times 27 = 0 \)
(xvii) \( (2a^3)^4 (4a^2)^2 = 16a^{12} \times 16a^4 = 256a^{16} \)
(xviii) \( (4x^2 y^3)^3 \div (3x^2 y^3)^3 = \frac{64x^6 y^9}{27x^6 y^9} = \frac{64}{27} \)
(xix) \( \left(\frac{1}{2x}\right)^3 \times (6x)^2 = \frac{1}{8x^3} \times 36x^2 = \frac{36x^2}{8x^3} = \frac{9}{2x} \)
(xx) \( \left(\frac{1}{4ab^2c}\right)^2 \div \left(\frac{3}{2a^2bc^2}\right)^4 = \frac{1}{16a^2 b^4 c^2} \times \frac{16a^8 b^4 c^8}{81} = \frac{a^6 c^6}{81} \)
(xxi) \( \frac{(5x^7)^3 \cdot (10x^2)^2}{(2x^6)^7} = \frac{125x^{21} \cdot 100x^4}{128x^{42}} = \frac{12500x^{25}}{128x^{42}} = \frac{3125}{32x^{17}} \)
(xxii) \( \frac{(7p^2 q^9 r^5)^2 (4pqr)^3}{(14p^6 q^{10} r^4)^2} = \frac{49p^4 q^{18} r^{10} \cdot 64p^3 q^3 r^3}{196p^{12} q^{20} r^8} = \frac{3136 p^7 q^{21} r^{13}}{196 p^{12} q^{20} r^8} = 16 p^{7-12} q^{21-20} r^{13-8} = 16p^{-5}qr^5 = \frac{16qr^5}{p^5} \br />In simple words: When simplifying, multiply the numbers together first. Then combine the same letters by adding their powers if they are multiplied, or subtracting if they are divided. Keep the final powers positive.

Exam Tip: Be very careful with negative bases. Raising a negative number to an even power makes it positive, while an odd power keeps it negative.

Question 4. Simplify and express the Solution in the positive exponent form :
(i) \( \frac{(-3)^3 \times 2^6}{6 \times 2^3} \)
(ii) \( \frac{(2^3)^5 \times 5^4}{4^3 \times 5^2} \)
(iii) \( \frac{36 \times (-6)^2 \times 3^6}{12^3 \times 3^5} \)
(iv) \( -\frac{128}{2187} \)
(v) \( \frac{a^{-7} \times b^{-7} \times c^5 \times d^4}{a^3 \times b^{-5} \times c^{-3} \times d^8} \)
(vi) \( (a^3 b^{-5})^{-2} \)
Answer:
(i) \( \frac{(-3)^3 \times 2^6}{6 \times 2^3} = \frac{-3^3 \times 2^6}{2 \times 3 \times 2^3} = \frac{-3^3 \times 2^6}{3 \times 2^4} = -3^{3-1} \times 2^{6-4} = -3^2 \times 2^2 = -6^2 \)
(ii) \( \frac{(2^3)^5 \times 5^4}{4^3 \times 5^2} = \frac{2^{15} \times 5^4}{(2^2)^3 \times 5^2} = \frac{2^{15} \times 5^4}{2^6 \times 5^2} = 2^{15-6} \times 5^{4-2} = 2^9 \times 5^2 \)
(iii) \( \frac{36 \times (-6)^2 \times 3^6}{12^3 \times 3^5} = \frac{6^2 \times 6^2 \times 3^6}{(2^2 \times 3)^3 \times 3^5} = \frac{6^4 \times 3^6}{2^6 \times 3^3 \times 3^5} = \frac{2^4 \times 3^4 \times 3^6}{2^6 \times 3^8} = \frac{2^4 \times 3^{10}}{2^6 \times 3^8} = \frac{3^2}{2^2} = \left(\frac{3}{2}\right)^2 \)
(iv) \( -\frac{128}{2187} = -\frac{2^7}{3^7} = -\left(\frac{2}{3}\right)^7 \)
(v) \( \frac{a^{-7} \times b^{-7} \times c^5 \times d^4}{a^3 \times b^{-5} \times c^{-3} \times d^8} = a^{-7-3} \cdot b^{-7-(-5)} \cdot c^{5-(-3)} \cdot d^{4-8} = a^{-10} b^{-2} c^8 d^{-4} = \frac{c^8}{a^{10} b^2 d^4} \)
(vi) \( (a^3 b^{-5})^{-2} = a^{3 \times (-2)} b^{-5 \times (-2)} = a^{-6} b^{10} = \frac{b^{10}}{a^6} \)
In simple words: Break down large numbers into prime factors first. Then apply exponent rules to simplify the powers and move negative power terms to make them positive.

Exam Tip: Always convert bases like \( 4, 6, 12, 36 \) into prime factors like \( 2 \) and \( 3 \) to make division and multiplication simpler.

Question 5. Evaluate
(i) \( 6^{-2} \div (4^{-2} \times 3^{-2}) \)
(ii) \( \left[\left(\frac{5}{6}\right)^2 \times \frac{9}{4}\right] \div \left[\left(-\frac{3}{2}\right)^2 \times \frac{125}{216}\right] \)
(iii) \( 5^3 \times 3^2 + (17)^0 \times 7^3 \)
(iv) \( 2^5 \times 15^0 + (-3)^3 - \left(\frac{2}{7}\right)^{-2} \)
(v) \( (2^2)^0 + 2^{-4} \div 2^{-6} + \left(\frac{1}{2}\right)^{-3} \)
(vi) \( 5^n \times 25^{n-1} \div (5^{n-1} \times 25^{n-1}) \)
Answer:
(i) \( 6^{-2} \div (4^{-2} \times 3^{-2}) = \left(\frac{1}{6}\right)^2 \div \left(\left(\frac{1}{4}\right)^2 \times \left(\frac{1}{3}\right)^2\right) = \frac{1}{36} \div \left(\frac{1}{16} \times \frac{1}{9}\right) = \frac{1}{36} \div \frac{1}{144} = \frac{1}{36} \times 144 = 4 \)
(ii) \( \left[\left(\frac{5}{6}\right)^2 \times \frac{9}{4}\right] \div \left[\left(-\frac{3}{2}\right)^2 \times \frac{125}{216}\right] = \left[\frac{25}{36} \times \frac{9}{4}\right] \div \left[\frac{9}{4} \times \frac{125}{216}\right] = \left[\frac{25}{16}\right] \div \left[\frac{125}{96}\right] = \frac{25}{16} \times \frac{96}{125} = \frac{6}{5} = 1\frac{1}{5} \)
(iii) \( 5^3 \times 3^2 + (17)^0 \times 7^3 = 125 \times 9 + 1 \times 343 = 1125 + 343 = 1468 \)
(iv) \( 2^5 \times 15^0 + (-3)^3 - \left(\frac{2}{7}\right)^{-2} = 32 \times 1 + (-27) - \left(\frac{7}{2}\right)^2 = 32 - 27 - \frac{49}{4} = 5 - \frac{49}{4} = \frac{20-49}{4} = -\frac{29}{4} = -7\frac{1}{4} \)
(v) \( (2^2)^0 + 2^{-4} \div 2^{-6} + \left(\frac{1}{2}\right)^{-3} = 1 + 2^{-4 - (-6)} + 2^3 = 1 + 2^2 + 8 = 1 + 4 + 8 = 13 \)
(vi) \( 5^n \times 25^{n-1} \div (5^{n-1} \times 25^{n-1}) = \frac{5^n \times 25^{n-1}}{5^{n-1} \times 25^{n-1}} = \frac{5^n}{5^{n-1}} = 5^{n-(n-1)} = 5^1 = 5 \)
In simple words: Solve the numbers inside the brackets first. Convert negative powers to positive by flipping fractions, and remember that any number with a power of 0 is equal to 1.

Exam Tip: Be careful with signs. A negative number squared, like \( (-\frac{3}{2})^2 \), becomes positive \( \frac{9}{4} \).

Question 6. If m = -2 and n = 2; find the values of:
(i) \( m^2 + n^2 - 2mn \)
(ii) \( m^n + n^m \)
(iii) \( 6m^{-3} + 4n^2 \)
(iv) \( 2n^3 - 3m \)
Answer:
Given: \( m = -2 \) and \( n = 2 \)
(i) \( m^2 + n^2 - 2mn = (-2)^2 + (2)^2 - 2(-2)(2) = 4 + 4 - (-8) = 8 + 8 = 16 \)
(ii) \( m^n + n^m = (-2)^2 + (2)^{-2} = 4 + \frac{1}{2^2} = 4 + \frac{1}{4} = \frac{17}{4} = 4\frac{1}{4} \)
(iii) \( 6m^{-3} + 4n^2 = 6(-2)^{-3} + 4(2)^2 = 6\left(\frac{1}{(-2)^3}\right) + 4(4) = 6\left(-\frac{1}{8}\right) + 16 = -\frac{3}{4} + 16 = \frac{-3+64}{4} = \frac{61}{4} = 15\frac{1}{4} \)
(iv) \( 2n^3 - 3m = 2(2)^3 - 3(-2) = 2(8) - (-6) = 16 + 6 = 22 \)
In simple words: Replace the letters with the given numbers. Be very careful with minus signs when multiplying or raising to a power.

Exam Tip: When substituting negative values like \( m = -2 \), always enclose them in brackets to prevent sign errors during calculation.