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Indices - Laws of Exponents

Key Concepts

Solved Examples: 41

Practice Questions: 3

Solved Examples

Question 1: Find the value of: 5³ × 5⁰ + 2⁴

We know: 5⁰ = 1
Therefore,
5³ × 5⁰ + 2⁴
= 5³ × 1 + 16
= 125 + 16
= 141

Question 2: Find the value of (3⁴ × 9²) ÷ 27²

Convert all terms into the same base:
9 = 3²
27 = 3³
Therefore,
(3⁴ × (3²)²) ÷ (3³)²
= (3⁴ × 3⁴) ÷ 3⁶
Using laws of indices:
= 3⁸ ÷ 3⁶
= 3⁸⁻⁶
= 3²
= 9

Question 3: 31⁷·⁵ ÷ 31³⁄² × 31⁻³ = (√31)ˣ

31⁷·⁵ ÷ 31³⁄² × 31⁻³ = (√31)ˣ
or, 31⁷·⁵ ÷ 31¹·⁵ × 31⁻³ = (√31)ˣ
or, 31⁷·⁵ ⁻ ¹·⁵ ⁻ ³ = (√31)ˣ
or, 31³ = (√31)ˣ
or, 31³ = 31ˣ⁄²
Since the bases are equal
x/2 = 3,
or x = 6

Question 4: If 2ᵃ = 8ᵇ and a + b = 15, find a.

Given 2ᵃ = 8ᵇ
or, 2ᵃ = (2³)ᵇ
2ᵃ = 2³ᵇ
∴ a = 3b
Also, a + b = 15
Substituting a = 3b we get
3b + b = 15
or, 4b = 15
or, b = 15⁄4
∴ a = 3 × 15⁄4
= 45⁄4

Question 5: Which of the following is larger 2³⁰⁰ or 3²⁰⁰?

Rewrite both expressions with the same power:
2³⁰⁰ = (2³)¹⁰⁰
3²⁰⁰ = (3²)¹⁰⁰
Now compare the bases:
2³ = 8
3² = 9
Therefore,
(2³)¹⁰⁰ = 8¹⁰⁰
(3²)¹⁰⁰ = 9¹⁰⁰
Since, 9¹⁰⁰ > 8¹⁰⁰
Therefore, 3²⁰⁰ > 2³⁰⁰

Question 6: Find the value of x from the equation: 2ˣ × 8¹⁄⁴ = 2¹⁄⁴

Convert 8 into a power of 2: 8 = 2³
Therefore,
8¹⁄⁴ = (2³)¹⁄⁴
= 2³⁄⁴
Now the equation becomes:
2ˣ × 2³⁄⁴ = 2¹⁄⁴
Therefore,
2ˣ⁺³⁄⁴ = 2¹⁄⁴
Since the bases are the same:
x + 3⁄4 = 1⁄4
x = 1⁄4 − 3⁄4
x = −2⁄4
x = −1⁄2

Question 7: If 9ˣ − 9ˣ⁻¹ = 648 find the value of xˣ

Given,
9ˣ − 9ˣ⁻¹ = 648
Taking 9ˣ⁻¹ common:
9ˣ⁻¹(9 − 1) = 648
9ˣ⁻¹ × 8 = 648
9ˣ⁻¹ = 648 ÷ 8
9ˣ⁻¹ = 81
Since,
81 = 9²
Therefore,
9ˣ⁻¹ = 9²
Comparing the powers:
x − 1 = 2
x = 3
Now,
xˣ = 3³
= 27

Question 8: If 4ˣ⁻ʸ = 64 and 4ˣ⁺ʸ = 1024 find the value of x.

Convert the numbers into powers of 4:
64 = 4³
1024 = 4⁵
Therefore, 4ˣ⁻ʸ = 4³
Comparing the powers: x − y = 3 … (1)
Also,
4ˣ⁺ʸ = 4⁵
Comparing the powers: x + y = 5 … (2)
Adding equations (1) and (2):
(x − y) + (x + y) = 3 + 5
2x = 8
x = 4

Question 9: If 2²ⁿ⁻¹ = 1/8ⁿ⁻³ find the value of n.

Given: 2²ⁿ⁻¹ = 1/8ⁿ⁻³
Convert 8 into a power of 2: 8 = 2³
Therefore,
1/8ⁿ⁻³ = 1/(2³)ⁿ⁻³
= 1/2³ⁿ⁻⁹
= 2⁻³ⁿ⁺⁹
Now the equation becomes:
2²ⁿ⁻¹ = 2⁻³ⁿ⁺⁹
Since the bases are the same, equate the powers:
2n − 1 = −3n + 9
5n = 10
n = 2

Question 10: If 2ⁿ⁺⁴ − 2ⁿ⁺² = 3 find the value of n.

Given: 2ⁿ⁺⁴ − 2ⁿ⁺² = 3
Taking 2ⁿ⁺² common:
2ⁿ⁺²(2² − 1) = 3
2ⁿ⁺²(4 − 1) = 3
3 × 2ⁿ⁺² = 3
Dividing both sides by 3:
2ⁿ⁺² = 1
Since, 2⁰ = 1
Therefore,
n + 2 = 0
n = −2

Question 11: Simplify [(2³)⁴ × (2²)⁵] ÷ (2⁵)²

Apply the power-of-a-power rule
(aᵐ)ⁿ = aᵐⁿ
= [2¹² × 2¹⁰] ÷ 2¹⁰
= 2¹²⁺¹⁰⁻¹⁰
= 2¹²

Question 12: If [(5²)ˣ]³ = 5²⁴ find x.

[(5²)ˣ]³
= 5²ˣ×³
= 5⁶ˣ
Hence,
5⁶ˣ = 5²⁴
∴ 6x = 24
or, x = 4

Question 13: Simplify [(3² × 5³)² × (3³ × 5)³] ÷ (3⁵ × 5⁴)²

Expand each power
= [(3⁴ × 5⁶)(3⁹ × 5³)] ÷ (3¹⁰ × 5⁸)
= (3¹³ × 5⁹) ÷ (3¹⁰ × 5⁸)
= 3³ × 5

Question 14: If [(2x⁻¹)³ ÷ (4x⁻²)²] = 2⁻¹⁄x find x, where x ≠ 0.

Expand the powers
(2x⁻¹)³ = 2³x⁻³
and
(4x⁻²)² = 2⁴x⁻⁴
Hence,
2³x⁻³ ÷ 2⁴x⁻⁴
= 2³⁻⁴ × x⁻³⁻⁽⁻⁴⁾
= 2⁻¹ × x
= x⁄2
Thus, x⁄2 = 1⁄(2x)
Multiplying by 2x
x² = 1
∴ x = ±1

Question 15: If 2⁻ˣ + 2⁻ˣ⁻¹ = 3⁄16 find x.

2⁻ˣ⁻¹ = 1⁄2 × 2⁻ˣ
Hence, 2⁻ˣ + ½(2⁻ˣ) = 3⁄16
3⁄2 × 2⁻ˣ = 3⁄16
∴ 2⁻ˣ = 1⁄8
= 2⁻³
Hence,
−x = −3
∴ x = 3

Question 16: Simplify (81¹⁄² + 32¹⁄⁵)²

81¹⁄² = √81 = 9
and
32¹⁄⁵ = ⁵√32 = 2
Hence,
(81¹⁄² + 32¹⁄⁵)²
= (9 + 2)²
= 11²
= 121

Question 17: Which is greater? (2³ × 3²)⁴ or (2² × 3³)³

Expand the powers
(2³ × 3²)⁴
= 2¹² × 3⁸
and
(2² × 3³)³
= 2⁶ × 3⁹
Divide the first by the second:
(2¹² × 3⁸) ÷ (2⁶ × 3⁹)
= 2⁶ ÷ 3
= 64⁄3
Since 64⁄3 > 1,
∴ (2³ × 3²)⁴ > (2² × 3³)³

Question 18: If 4ˣ⁺¹ × 8²ˣ⁻¹ = 2²⁰ find x.

Given 4ˣ⁺¹ × 8²ˣ⁻¹ = 2²⁰
Express 4 and 8 as powers of 2
(2²)ˣ⁺¹ × (2³)²ˣ⁻¹ = 2²⁰
or, 2²ˣ⁺² × 2⁶ˣ⁻³ = 2²⁰
or, 2⁸ˣ⁻¹ = 2²⁰
∴ 8x − 1 = 20
or, 8x = 21
∴ x = 21⁄8

Question 19: If 9ˣ⁻¹ ÷ 27ˣ⁻² = 3⁵ find x.

9ˣ⁻¹ ÷ 27ˣ⁻² = 3⁵
Express 9 and 27 as powers of 3
(3²)ˣ⁻¹ ÷ (3³)ˣ⁻² = 3⁵
= 3²ˣ⁻² ÷ 3³ˣ⁻⁶
= 3²ˣ⁻²⁻³ˣ⁺⁶
= 3⁻ˣ⁺⁴
Hence,
−x + 4 = 5
∴ x = −1

Question 20: If 2ˣ × 4ˣ⁺¹ = 8²ˣ⁻¹ find x.

2ˣ × 4ˣ⁺¹ = 8²ˣ⁻¹
Express 4 and 8 as powers of 2
2ˣ × (2²)ˣ⁺¹ = (2³)²ˣ⁻¹
or, 2ˣ × 2²ˣ⁺² = 2⁶ˣ⁻³
or, 2³ˣ⁺² = 2⁶ˣ⁻³
∴ 3x + 2 = 6x − 3
or, 3x = 5
or, x = 5⁄3

Question 21: Simplify [(a³b⁻²)² × (a⁻¹b³)³] ÷ (a²b)⁻¹

[(a³b⁻²)² × (a⁻¹b³)³] ÷ (a²b)⁻¹
Expand the powers:
= [a⁶b⁻⁴ × a⁻³b⁹] ÷ a⁻²b⁻¹
= a³b⁵ ÷ a⁻²b⁻¹
= a³⁻⁽⁻²⁾b⁵⁻⁽⁻¹⁾
= a⁵b⁶

Question 22: Simplify [(x²y⁻³)³ × (x⁻¹y²)²]⁻¹

[(x²y⁻³)³ × (x⁻¹y²)²]⁻¹
First simplify the expression inside the brackets
= [x⁶y⁻⁹ × x⁻²y⁴]⁻¹
= [x⁴y⁻⁵]⁻¹
Applying the power −1
= x⁻⁴y⁵
= y⁵⁄x⁴

Question 23: Simplify [(a²b³)² ÷ (a⁻¹b)³] × (a⁻²b⁴)⁻¹

Expand the powers
= [a⁴b⁶ ÷ a⁻³b³] × a²b⁻⁴
= a⁴⁻⁽⁻³⁾b⁶⁻³ × a²b⁻⁴
= a⁷b³ × a²b⁻⁴
= a⁹b⁻¹

Question 24: Evaluate (16³⁄⁴ − 81¹⁄²)² + 125²⁄³

Now 16³⁄⁴ = (⁴√16)³ = 2³ = 8
and 81¹⁄² = √81 = 9
also 125²⁄³ = (∛125)² = 5² = 25
Hence, from above
(16³⁄⁴ − 81¹⁄²)² + 125²⁄³
= (8 − 9)² + 25
= (−1)² + 25
= 1 + 25
= 26

Question 25: If (2ˣ⁺¹ × 4²ˣ⁻¹) ÷ 8ˣ⁻² = 16³ find x.

Given (2ˣ⁺¹ × 4²ˣ⁻¹) ÷ 8ˣ⁻² = 16³
Express everything as powers of 2
[2ˣ⁺¹ × (2²)²ˣ⁻¹] ÷ (2³)ˣ⁻² = (2⁴)³
= [2ˣ⁺¹ × 2⁴ˣ⁻²] ÷ 2³ˣ⁻⁶
= 2ˣ⁺¹⁺⁴ˣ⁻²⁻³ˣ⁺⁶
= 2²ˣ⁺⁵
Hence,
2²ˣ⁺⁵ = 2¹²
∴ 2x + 5 = 12
2x = 7
or, x = 7⁄2

Question 26: If 2ˣ⁺² − 2ˣ = 48 find x.

Factor out 2ˣ:
2ˣ(2² − 1) = 48
3 × 2ˣ = 48
2ˣ = 16
= 2⁴
∴ x = 4

Question 27: If 2ˣ + 2⁻ˣ = 5 find 4ˣ + 4⁻ˣ

Given 2ˣ + 2⁻ˣ = 5
Square both sides
(2ˣ + 2⁻ˣ)² = 25
or, 2²ˣ + 2 + 2⁻²ˣ = 25
Hence,
2²ˣ + 2⁻²ˣ = 23
Since 2²ˣ = 4ˣ and 2⁻²ˣ = 4⁻ˣ,
∴ 4ˣ + 4⁻ˣ = 23

Question 28: If 3ˣ + 3⁻ˣ = 10 find 27ˣ + 27⁻ˣ

Let a = 3ˣ + 3⁻ˣ = 10.
Using (a + b)³ = a³ + b³ + 3ab(a + b),
where a = 3ˣ, b = 3⁻ˣ and ab = 1,
we get 10³ = 3³ˣ + 3⁻³ˣ + 3(1)(10)
or, 1000 = 27ˣ + 27⁻ˣ + 30
∴ 27ˣ + 27⁻ˣ = 970

Question 29: If 2ˣ + 2⁻ˣ = 6 find 8ˣ + 8⁻ˣ

Let a = 2ˣ
Then a + 1⁄a = 6
We need to find a³ + 1⁄a³
Using a³ + 1⁄a³ = (a + 1⁄a)³ − 3(a + 1⁄a),
we get
= 6³ − 3 × 6
= 216 − 18
= 198

Question 30: If 2ˣ + 2⁻ˣ = 3 find 2²ˣ + 2⁻²ˣ + 2³ˣ + 2⁻³ˣ

2ˣ + 2⁻ˣ = 3
Squaring both sides
(2ˣ + 2⁻ˣ)² = 3²
or, 2²ˣ + 2 + 2⁻²ˣ = 9
∴ 2²ˣ + 2⁻²ˣ = 7
Now, (2ˣ + 2⁻ˣ)³
= 2³ˣ + 2⁻³ˣ + 3(2ˣ + 2⁻ˣ)
Hence,
2³ˣ + 2⁻³ˣ = 3³ − 3 × 3
= 27 − 9
= 18
Therefore, the required value is
7 + 18 = 25

Question 31: If x¹⁄² + x⁻¹⁄² = 4 find x³⁄² + x⁻³⁄² where x > 0.

Let a = x¹⁄².
Then a + 1⁄a = 4.
We know
a³ + 1⁄a³ = (a + 1⁄a)³ − 3(a + 1⁄a),
we get:
a³ + 1⁄a³ = 4³ − 3 × 4
= 64 − 12
= 52

Question 32: Find the positive real number solution for x that satisfies the equation: (x − 2)⁵ = 32

Given (x − 2)⁵ = 32
Since 32 = 2⁵,
we have
(x − 2)⁵ = 2⁵
∴ x − 2 = 2
or, x = 4
Since 4 is positive, it satisfies the condition.

Question 33: Solve the exponential equation containing a fractional inverse root format: (2/3)⁽³ˣ ⁻ ⁴⁾ = (27/8)⁽ˣ ⁻ ¹⁾

Given(2⁄3)³ˣ⁻⁴ = (27⁄8)ˣ⁻¹
Rewrite the right-hand side
27⁄8 = (3⁄2)³ = (2⁄3)⁻³
Therefore,
(2⁄3)³ˣ⁻⁴ = [(2⁄3)⁻³]ˣ⁻¹
= (2⁄3)⁻³ˣ⁺³
Since the bases are equal:
3x − 4 = −3x + 3
6x = 7
∴ x = 7⁄6

Question 34: Solve for x in the exponential equation: 4ˣ − 6 × 2ˣ + 8 = 0

Given 4ˣ − 6 × 2ˣ + 8 = 0
Since 4ˣ = (2²)ˣ = (2ˣ)²,
let 2ˣ = y.
Then y² − 6y + 8 = 0
or, (y − 2)(y − 4) = 0
∴ y = 2 or y = 4
Therefore 2ˣ = 2 ⟹ x = 1
or, 2ˣ = 4 = 2² ⟹ x = 2
∴ x = 1 or x = 2

Question 35: Find the real number value of x that satisfies the equation: 3⁽ˣ ⁺ ¹⁾ + 3⁽² ⁻ ˣ⁾ = 28

Given 3ˣ⁺¹ + 3²⁻ˣ = 28
Rewrite 3ˣ⁺¹ = 3 × 3ˣ
and 3²⁻ˣ = 9⁄3ˣ
Let 3ˣ = y
Then 3y + 9⁄y = 28
Multiplying by y
3y² + 9 = 28y
or, 3y² − 28y + 9 = 0
or, (3y − 1)(y − 9) = 0
or, y = 1⁄3 or y = 9
Since y = 3ˣ:
3ˣ = 3⁻¹ ⟹ x = −1
or 3ˣ = 3² ⟹ x = 2
∴ x = −1 or x = 2

Question 36: Find the values of x that satisfy the radical inverse exponential equation: 2ˣ + 2⁻ˣ = 2.5

Given 2ˣ + 2⁻ˣ = 2.5
Let 2ˣ = y
Then 2⁻ˣ = 1⁄y
So, y + 1⁄y = 2.5
or, y² + 1 = 2.5y
or, 2y² − 5y + 2 = 0
or, (2y − 1)(y − 2) = 0
∴ y = 1⁄2 or y = 2
Since y = 2ˣ:
2ˣ = 2⁻¹ ⟹ x = −1
or 2ˣ = 2¹ ⟹ x = 1
∴ x = −1 or x = 1

Question 37: Solve the multi-stage exponential equation for x: 6 × 9ˣ − 13 × 6ˣ + 6 × 4ˣ = 0

Given 6 × 9ˣ − 13 × 6ˣ + 6 × 4ˣ = 0
Divide throughout by 6 × 4ˣ:
(9⁄4)ˣ − (13⁄6)(3⁄2)ˣ + 1 = 0
Since (9⁄4)ˣ = [(3⁄2)²]ˣ = [(3⁄2)ˣ]²,
let y = (3⁄2)ˣ
Then y² − (13⁄6)y + 1 = 0
or, 6y² − 13y + 6 = 0
or, (3y − 2)(2y − 3) = 0
∴ y = 2⁄3 or y = 3⁄2
Since y = (3⁄2)ˣ:
(3⁄2)ˣ = (3⁄2)⁻¹ ⟹ x = −1
or, (3⁄2)ˣ = (3⁄2)¹ ⟹ x = 1
∴ x = −1 or x = 1

Question 38: Find all positive integer solutions for x and y that satisfy the system of index variable equations: 2ˣ = 4ʸ and 9ˣ × 3ʸ = 243

Given 2ˣ = 4ʸ and 9ˣ × 3ʸ = 243
Solving the first equation
2ˣ = 4ʸ
or, 2ˣ = (2²)ʸ
or, 2ˣ = 2²ʸ
∴ x = 2y
Solving the second equation
9ˣ × 3ʸ = 243
or, (3²)ˣ × 3ʸ = 3⁵
or, 3²ˣ⁺ʸ = 3⁵
or, 2x + y = 5
Substitute x = 2y
2(2y) + y = 5
5y = 5
∴ y = 1
Therefore x = 2y = 2
∴ x = 2, y = 1

Question 39: If 9ˣ⁺² = 720 + 9ˣ, find the value of (4x)¹⁄ˣ.

9ˣ⁺² = 720 + 9ˣ
Using 9ˣ⁺² = 9² × 9ˣ:
81 × 9ˣ = 720 + 9ˣ
or, 81 × 9ˣ − 9ˣ = 720
or, 80 × 9ˣ = 720
or, 9ˣ = 9
∴ x = 1
Now,
(4x)¹⁄ˣ = (4 × 1)¹⁄¹
= 4¹
= 4

Question 40: If 49,392 = a⁴ × b² × c³, find the values of a, b and c, where a, b and c are different positive primes.

Factorise 49392
49392 = 2⁴ × 3³ × 7²
Comparing with a⁴ × b² × c³
we get
a = 2
b = 3
c = 7

Question 41: If a = xyᵖ⁻¹, b = xyᑫ⁻¹ and c = xyʳ⁻¹, prove that aᑫ⁻ʳ × bʳ⁻ᵖ × cᵖ⁻ᑫ = 1.

Given
a = xyᵖ⁻¹
b = xyᑫ⁻¹
c = xyʳ⁻¹
Substituting these values
aᑫ⁻ʳ × bʳ⁻ᵖ × cᵖ⁻ᑫ
= (xyᵖ⁻¹)ᑫ⁻ʳ × (xyᑫ⁻¹)ʳ⁻ᵖ × (xyʳ⁻¹)ᵖ⁻ᑫ
Using (ab)ⁿ = aⁿbⁿ:
= xᑫ⁻ʳy⁽ᵖ⁻¹⁾⁽ᑫ⁻ʳ⁾ × xʳ⁻ᵖy⁽ᑫ⁻¹⁾⁽ʳ⁻ᵖ⁾ × xᵖ⁻ᑫy⁽ʳ⁻¹⁾⁽ᵖ⁻ᑫ⁾
Collecting the powers of x:
= x⁽ᑫ⁻ʳ⁾⁺⁽ʳ⁻ᵖ⁾⁺⁽ᵖ⁻ᑫ⁾ × yᴱ
where
E = (p − 1)(q − r) + (q − 1)(r − p) + (r − 1)(p − q)
The exponent of x is:
(q − r) + (r − p) + (p − q) = 0
Now simplify E:
E = pq − pr − q + r + qr − pq − r + p + rp − qr − p + q
= 0
Hence,
aᑫ⁻ʳ × bʳ⁻ᵖ × cᵖ⁻ᑫ
= x⁰ × y⁰
= 1 × 1
= 1
∴ aᑫ⁻ʳ × bʳ⁻ᵖ × cᵖ⁻ᑫ = 1

Question 42: 4⁽⁶⁺¹²⁺¹⁸⁺²⁴⁺⋯⁺⁶ˣ⁾ = (0.0625)⁻⁸⁴. Find the value of x.

Given 4⁽⁶⁺¹²⁺¹⁸⁺²⁴⁺⋯⁺⁶ˣ⁾ = (0.0625)⁻⁸⁴
Since 0.0625 = 1/16 = 4⁻²,
we get
(0.0625)⁻⁸⁴ = (4⁻²)⁻⁸⁴ = 4¹⁶⁸.
Hence,
4⁽⁶⁺¹²⁺¹⁸⁺²⁴⁺⋯⁺⁶ˣ⁾ = 4¹⁶⁸
∴ 6 + 12 + 18 + 24 + ⋯ + 6x = 168
Factor out 6:
6(1 + 2 + 3 + ⋯ + x) = 168
Using 1 + 2 + 3 + ⋯ + x = x(x + 1)/2,
we get
6 × x(x + 1)/2 = 168
or, 3x(x + 1) = 168
or, x(x + 1) = 56
or, x² + x − 56 = 0
or, (x + 8)(x − 7) = 0
Since x is positive,
∴ x = 7.

Question 43: If a, b and c are non-zero and 14ᵃ = 36ᵇ = 84ᶜ, then find the value of 6b(1/c − 1/a).

Let 14ᵃ = 36ᵇ = 84ᶜ = k.
Taking logarithms:
a = log k / log 14,
b = log k / log 36,
c = log k / log 84.
Now,
1/c − 1/a = log 84 / log k − log 14 / log k
= (log 84 − log 14) / log k
= log(84/14) / log k
= log 6 / log k.
Therefore,
6b(1/c − 1/a)
= 6 × (log k / log 36) × (log 6 / log k)
= 6 log 6 / log 36.
Since 36 = 6²,
log 36 = 2 log 6.
Hence, 6 log 6 / (2 log 6)
= 3

Question 44: If x² + x + 1 = 0, then x²⁰¹⁸ + x²⁰¹⁹ equals which of the following?

We know that
x³ − 1 = (x − 1)(x² + x + 1)
Since x² + x + 1 = 0
∴ x³ − 1 = 0
or, x³ = 1
Now, x²⁰¹⁸ + x²⁰¹⁹
= (x³)⁶⁷² × x² + (x³)⁶⁷³
= 1⁶⁷² × x² + 1⁶⁷³
= x² + 1
From x² + x + 1 = 0,
x² + 1 = −x
∴ x²⁰¹⁸ + x²⁰¹⁹ = −x

Question 45: If 9²ˣ⁻¹ − 81ˣ⁻¹ = 1944, then find the value of x.

We have 9²ˣ⁻¹ − 81ˣ⁻¹ = 1944
Since 81 = 9²,
9²ˣ⁻¹ − 9²ˣ⁻² = 1944
Factor out 9²ˣ⁻²:
9²ˣ⁻²(9 − 1) = 1944
or, 9²ˣ⁻² × 8 = 1944
∴ 9²ˣ⁻² = 243
Since 9 = 3² and 243 = 3⁵,
(3²)²ˣ⁻² = 3⁵
or, 3⁴ˣ⁻⁴ = 3⁵
∴ 4x − 4 = 5
or, 4x = 9
∴ x = 9⁄4

Question 46: If 5ˣ − 3ʸ = 13438 and 5ˣ⁻¹ + 3ʸ⁺¹ = 9686, then find the value of x + y.

We have
5ˣ − 3ʸ = 13438 …(1)
5ˣ⁻¹ + 3ʸ⁺¹ = 9686 …(2)
Since 5ˣ = 5 × 5ˣ⁻¹
and 3ʸ⁺¹ = 3 × 3ʸ,
let 5ˣ⁻¹ = a and 3ʸ = b.
Then equations (1) and (2) become
5a − b = 13438 …(3)
a + 3b = 9686 …(4)
From (3),
b = 5a − 13438
Substituting in (4),
a + 3(5a − 13438) = 9686
or, a + 15a − 40314 = 9686
or, 16a = 50000
or, a = 3125
∴ 5ˣ⁻¹ = 3125 = 5⁵
or, x − 1 = 5
∴ x = 6
Now, from (3),
5(3125) − b = 13438
15625 − b = 13438
b = 2187
∴ 3ʸ = 2187 = 3⁷
or, y = 7
Hence,
x + y = 6 + 7
∴ x + y = 13

Question 47: If 9ᵃ²⁺²ᵃ⁻³ − 4 × 3ᵃ²⁺²ᵃ⁻² + 27 = 0, then what is the product of all possible values of a?

Let 3ᵃ²⁺²ᵃ⁻³ = t
Then 9ᵃ²⁺²ᵃ⁻³ = t²
and 3ᵃ²⁺²ᵃ⁻² = 3t.
∴ t² − 4(3t) + 27 = 0
or, t² − 12t + 27 = 0
or, (t − 3)(t − 9) = 0
∴ t = 3 or t = 9
Case 1: t = 3
3ᵃ²⁺²ᵃ⁻³ = 3
∴ a² + 2a − 3 = 1
or, a² + 2a − 4 = 0
The product of the roots is −4.
Case 2: t = 9
3ᵃ²⁺²ᵃ⁻³ = 9 = 3²
∴ a² + 2a − 3 = 2
or, a² + 2a − 5 = 0
The product of the roots is −5.
Hence, the product of all possible values of a is
(−4)(−5) = 20

Question 48: If 12¹²ˣ × 4²⁴ˣ⁺¹² × 5²ʸ = 8⁴ᶻ × 20¹²ˣ × 243³ˣ⁻⁶, where x, y and z are natural numbers, then find the value of x + y + z.

Express all the bases in terms of their prime factors.
(2² × 3)¹²ˣ × (2²)²⁴ˣ⁺¹² × 5²ʸ
= (2³)⁴ᶻ × (2² × 5)¹²ˣ × (3⁵)³ˣ⁻⁶
Simplifying,
2²⁴ˣ × 3¹²ˣ × 2⁴⁸ˣ⁺²⁴ × 5²ʸ
= 2¹²ᶻ × 2²⁴ˣ × 5¹²ˣ × 3¹⁵ˣ⁻³⁰
Collecting like bases,
2⁷²ˣ⁺²⁴ × 3¹²ˣ × 5²ʸ
= 2¹²ᶻ⁺²⁴ˣ × 3¹⁵ˣ⁻³⁰ × 5¹²ˣ
Since the prime factorisations are equal, the exponents of each prime must be equal.
For 3:
12x = 15x − 30
3x = 30
∴ x = 10
For 5:
2y = 12x
2y = 120
∴ y = 60
For 2:
72x + 24 = 12z + 24x
48x + 24 = 12z
Substituting x = 10,
480 + 24 = 12z
504 = 12z
∴ z = 42
Hence,
x + y + z = 10 + 60 + 42
∴ x + y + z = 112

Question 49: If 3ᵃ = 4, 4ᵇ = 5, 5ᶜ = 6, 6ᵈ = 7, 7ᵉ = 8 and 8ᶠ = 9, then what is the value of the product abcdef?

Using the change-of-base formula,
a = log₃4 = log 4 / log 3
b = log₄5 = log 5 / log 4
c = log₅6 = log 6 / log 5
d = log₆7 = log 7 / log 6
e = log₇8 = log 8 / log 7
f = log₈9 = log 9 / log 8
Therefore, abcdef
= (log 4 / log 3) × (log 5 / log 4) × (log 6 / log 5) × (log 7 / log 6) × (log 8 / log 7) × (log 9 / log 8)
All the intermediate terms cancel out
abcdef = log 9 / log 3
= log₃9
= 2

Question 50: What is the sum of all real values of k for which (1⁄8)ᵏ × (1⁄32768)¹⁄³ = (1⁄8) × (1⁄32768)¹⁄ᵏ?

We know that 8 = 2³
and 32768 = 2¹⁵.
Therefore, 1⁄8 = 2⁻³
and 1⁄32768 = 2⁻¹⁵.
Substituting,
(2⁻³)ᵏ × (2⁻¹⁵)¹⁄³ = 2⁻³ × (2⁻¹⁵)¹⁄ᵏ
Using the laws of exponents,
2⁻³ᵏ × 2⁻⁵ = 2⁻³ × 2⁻¹⁵⁄ᵏ
∴ 2⁻³ᵏ⁻⁵ = 2⁻³⁻¹⁵⁄ᵏ
Since the bases are equal,
−3k − 5 = −3 − 15⁄k
or, −3k² − 5k = −3k − 15
or, 3k² + 2k − 15 = 0
For a quadratic equation ak² + bk + c = 0, the sum of the roots is −b⁄a.
Here, a = 3, b = 2
∴ Sum of all real values of k
= −2⁄3

Question 51: If (5.55)ˣ = (0.555)ʸ = 1000, then find the value of (1⁄x − 1⁄y)

Since (5.55)ˣ = 1000,
taking reciprocals of the exponents,
1⁄x = log₁₀(5.55).
Similarly,
(0.555)ʸ = 1000
gives 1⁄y = log₁₀(0.555).
Therefore,
1⁄x − 1⁄y
= log₁₀(5.55) − log₁₀(0.555)
= log₁₀(5.55⁄0.555)
= log₁₀(10)
∴ 1⁄x − 1⁄y = 1.

What are Indices?

An index (plural: indices) is a small number written slightly above and to the right of another number or variable. It indicates how many times the base is multiplied by itself. Indices provide a compact way to represent repeated multiplication and simplify lengthy mathematical expressions.
For example, 5³ means 5 × 5 × 5, where 5 is the base and 3 is the index (or exponent). Similarly, x⁴ means x × x × x × x.

Parts of an Exponential Expression

Every exponential expression consists of two parts: the base and the index (or exponent). The base is the number or variable that is multiplied repeatedly, while the index indicates how many times the base is multiplied by itself.
Examples:
• In 7⁴, the base is 7 and the index is 4.
• In x⁵, the base is x and the index is 5.
• In (ab)³, the base is ab and the index is 3.

Understanding Repeated Multiplication

An index represents repeated multiplication of the base by itself. Instead of writing the same factor multiple times, indices provide a shorter and more convenient form.
Examples:
2⁵ = 2 × 2 × 2 × 2 × 2 = 32
4³ = 4 × 4 × 4 = 64
x⁴ = x × x × x × x

Types of Indices

Indices can be positive, zero or negative. Each type follows a specific rule.
Positive Index: A positive index indicates repeated multiplication.
Example: 3⁴ = 3 × 3 × 3 × 3 = 81
Zero Index: Any non-zero number raised to the power 0 is equal to 1.
Examples: 8⁰ = 1, x⁰ = 1, where x ≠ 0
Negative Index: A negative index represents the reciprocal of the corresponding positive index.
Examples: 2⁻³ = 1/2³ = 1/8, x⁻² = 1/x²

Laws of Exponents

A number is divisible by 5 if its last digit is 0 or 5.
Examples: 85, 320, and 1,745 are divisible by 5 because they end in 0 or 5.

Divisibility Rule for 6

Indices follow a set of rules known as the laws of exponents, which simplify calculations involving powers.
The important laws are:
• aᵐ × aⁿ = aᵐ⁺ⁿ
• aᵐ ÷ aⁿ = aᵐ⁻ⁿ, where a ≠ 0
• (aᵐ)ⁿ = aᵐⁿ
• (ab)ⁿ = aⁿbⁿ
• (a/b)ⁿ = aⁿ/bⁿ, where b ≠ 0
• a⁰ = 1, where a ≠ 0
• a⁻ⁿ = 1/aⁿ, where a ≠ 0
These laws are the foundation for simplifying exponential expressions.

Summary of Indices - Laws of Exponents

Concept	Summary
Index (Exponent)	A small number written above and to the right of a base that indicates repeated multiplication.
Base	The number or variable that is repeatedly multiplied.
Positive Index	Represents repeated multiplication of the base.
Zero Index	For any non-zero base, a⁰ = 1.
Negative Index	Represents the reciprocal of the corresponding positive index. a⁻ⁿ = 1/aⁿ.
Product Rule	aᵐ × aⁿ = aᵐ⁺ⁿ
Quotient Rule	aᵐ ÷ aⁿ = aᵐ⁻ⁿ (a ≠ 0)
Power of a Power	(aᵐ)ⁿ = aᵐⁿ
Power of a Product	(ab)ⁿ = aⁿbⁿ
Power of a Quotient	(a/b)ⁿ = aⁿ/bⁿ (b ≠ 0)

Common Mistakes

Students often check the entire number instead of only the required digits. Example: For divisibility by 4, check only the last two digits, not the whole number.
Confusing the rules for 3 and 9. Some students add the digits correctly but forget that
1. The sum must be divisible by 3 for divisibility by 3.
2. The sum must be divisible by 9 for divisibility by 9.
Forgetting the rule for 6. A number is divisible by 6 only if it is divisible by both 2 and 3. Being divisible by just one of them is not enough.
Using the wrong digits for 8. For divisibility by 8, check only the last three digits. Checking the last two digits gives the wrong result.
Applying the rule for 11 incorrectly. When testing divisibility by 11, add the digits in alternate positions and find the difference between the two sums. Do not simply add all the digits together.

Practice Questions

Question: Find the greatest possible value of (a + b) for which the 8-digit number 143b203a is divisible by 15.

Question: If the number 59a44b is divisible by 36 then what is the maximum value of (a + b)?

Question: If a 5-digit number 535ab is divisible by 3, 7 and 11, then what is the value of (a – b)?