# Laws of Indices

**Further Mathematics SS 1 FIRST TERM**

**WEEK ONE**

**LAWS OF INDICES:**

**CONTENT:**

- Laws of Indices
- Application of Indices Linear Equations
- Application of Quadratic Equations

**LAWS OF INDICES**

There are laws governing the use of indices. These are useful in other subjects. They are;

**a**^{n }x a^{m }= a^{n+m}

Therefore

a^{3} x a^{3 }= a x a x a x a x a x a = a^{6 }i.e a^{3+3 } = a^{6}

In general, when multiplying indices with same base, you add the power

**a**^{n }÷ a^{m }= a^{n-m}.

a^{5 }÷ a^{4 }= a^{5-4 }= a^{1}

Also, a^{5 }÷ a^{4 }= = a

In general, when dividing indices with the same base, you subtract the power

**(a**^{n})^{m}= a^{n×m }= a^{mn}

(a^{3})^{2 }= a^{3×2 }= a^{6 }

**a**^{0}=1

a^{5 }÷ a^{5 }= a^{5-5 }= a^{0}

Also, a^{5 }÷ a^{5 }= = 1

This implies that anything to power zero is equal to 1, i. e 5^{0} = 1, 2^{0} = 1

**a**^{-n}= 1/a^{n}

Consider a^{5 }÷ a^{6 }= a^{5-6 }= a^{-1}

But a^{5}÷ a^{6 }= =

a^{-1 }=

In the same way a^{-2}=,

**a**^{ n/m }=^{n}= )

Consider (8)^{ 1/3} = (2^{3})^{ 1/3} = (2^{3})^{ 1/3 }= 2

= 2,

**Example1:**

(i) (a^{5 }x a^{6}) /a^{5} (ii) (2^{6} ÷2^{7} x 2^{4})^{1/3} (iii) 32^{1/5 }(iv) 216 ÷ 3^{4}

**Solution:**

- (a
^{5 }x a^{6}) /a^{5}= a^{3 +6 -5 }= a^{9-5}= a^{4} - (2
^{6}÷ 2^{7 }x 2^{4})^{1/3}= (2^{6-7+4})^{1/3}

Re arrange the indices

(2^{6+4-7})^{1/3 }= (2^{3})^{1/3 }= 2

iii. 32^{1/5} = 32^{1/5 }but 32 = 2^{5}

(2^{5})^{ 1/5} = 2^{5x 1/5 }= 2

iv. 216 ÷ 3^{4}

216 = 2^{3} x 3^{3}

2^{3} x 3^{3} ÷ 3^{4} = 2^{3} x 3^{3-4 }= 2^{3} x 3^{-1}

2^{3}/3 0r 8/3

**Example2:**

Simplify

**Solution**

= = =

**CLASS ACTIVITIES:**

- Evaluate each of the following

(a) 8^{0 }(b) 5^{-1} (c) 8^{2/3 }(d) (x^{3})^{-2/3 }(e) (4^{3})^{5}

- Evaluate each of the following

(f) (625)^{-1/4 }(g) 64^{2/3 }(h) 9^{1/3 }x 9^{1/6 }(i) 3^{6} ÷ 3^{7} x 2^{2} (j) (1000)^{-5/3}

**SUB TOPIC: APPLICATION OF THE LAWS OF INDICES.**

**Examples:**

- Solve (1/2)
^{x}= 8

**Solution:**

(1/2)^{x} = (2^{-1})^{x} = 2^{-x}

8 = 2^{3 }since we have same base, then –x = 3.

Multiply the equation by (-1)

X = -3

- Solve the equation 8
^{x}= 0.25

**Solution:**

8^{x} = (2^{3})^{x} = 2^{3x}

0.25 = 25/100 = ¼ = (1/2)^{2} = (2^{-1})^{2} = 2^{-2}

23x = 2-2

3x = -2

X = -2/3

- Solve for x in the equation: (0.25)
^{x+1}= 16

(1/4)^{x+1} 2^{4}

(2^{-2})^{x+1} = 2^{4}

2^{-2x-2} = 2^{4}

Equate the power

-2x – 2 = 4

-2x = 4 + 2 = 6

X = -6/2 = -3

X = -3

If 10^{-x} = 0.001. what is the value of x?

0.001 = 10^{-3}

–x = -3

x = 3

- If 25
^{(5x)}= 625, what is x.?

(5^{2})^{5x} = 5^{4}

5^{10x} = 5^{4}

10x = 4

X = 4/10 0r 2/5

**CLASS ACTIVITIES:**

Solve for x in the following equations

- 3
^{x }= 81 - 2
^{x }= 32 - 9
^{x}= 1/729 - 25
^{(5x) }= 625 - 2
^{x}x 4^{-x}= 2

**SUB TOPIC: APPLICATION OF INDICES LEADING TO QUADRATIC EQUATION**

Some exponential equations will lead to quadratic equations as you will see in the following examples.

- 5
^{2x}-30 x 5^{x}+ 125 =0

**Solution: **

Re-write the equation

(5^{x})^{2} -30 x 5^{x} + 125 =0

Let 5^{x }= p, then

P^{2} – 30p + 125 = 0

Solve for p by factorization

(p-5)(p-25) = 0

P – 5 = 0 or p – 25 = 0

Then p = 5 or 25,

Recall that p =5^{x }

Therefore 5^{x }= 5^{1}, then x = 1

or 5^{x }= 25 this means that 5^{x }= 5^{2}, x = 2

Solve the equation 2^{2x} + 4(^{2x}) – 32 = 0

2^{2x} + 4(^{2x}) – 32 = 0

(2^{2x})^{2} + 4(^{2x}) – 32 = 0

Let 2^{x }= y, then

y^{2 }+ 4y – 32 = 0

(y + 8)(y – 4) = 0

y = 4 or -8

Then 2^{x} = 2^{2} or 2^{x} = -8. But this (2^{x} = -8) has no solution

Therefore x = 2

- Solve for x in the equation

3^{2(x-1) }– 8(3^{(x-2)}) = 1

**Solution: **

Re write the equation

3^{2 }x 3^{-2 }– 8 x 3^{x }x 3^{-2 }– 1 = 0

3^{2x }x 1/3^{2 }– 8 x 3^{x }x 1/3^{2 }– 1 = 0

Multiply the equation by 3^{2}

3^{2x }– 8(3^{x}) – 3^{2 }= 0

Lep p = 3^{x}

P^{2} – 8p – 9 = 0

(p-9)(p+1) = 0

P = 9 or -1

Recall that p = 3^{x}

3^{x }= 3^{2}, 3^{x }= -1 has no solution

x = 2

**CLASS ACTIVITIES:**

Solve the following equations

- 2
^{2x }– 5(2^{x}) + 4 = 0 - 3
^{2x+1 }+ 26(3^{x}) – 9 = 0 - 2
^{2x }– 6(2^{x}) = -8 - 7
^{2x }– 2 x (7^{x}) = -1 - 2
^{x+3 }– 15 = 21^{1-x}

**PRACTICE EXERCISE:**

**Objective Test:**

Choose the correct answer from the options

- Simplify (105)
^{0}(a) 0 (b) 1 (c) 5 (d) 3 (e)-1 - Evaluate 343
^{2/3}(a) 7 (b) 49 (c) 343 (d) 3 (e) 9 - Simplify (2
^{8}x 4^{-3}) / 2^{6}(a) 1/16 (b) 16 (c) 2^{20 }(d) ¼ (e) 2/4 - Solve the equation 3
^{-x}= 243, x = ? (a) 5 (b) 3 (c) 4 (d) -5 (e) -3 - Solve the equation 3
^{2x }– 9 = 0, x = ? (a) ±3 (b) 2 (c) -2 (d) 1 (e) 0

**Essay questions: **

- Simplify (a)
^{4} - (1/3)
^{4}x 3^{6}÷ (2/3)^{2} - Solve for x, if 125
^{x-1 }= 25^{2x-3} - If 9
^{2x+1 }= 81^{x-2}/3^{x}, what is x? - Find the value of x satisfying
- 3
^{2x}– 30(3^{x}) + 81 = 0

**WEEKEND ASSIGNMENT:**

Simplify the following

- If =
- Solve
- If

**KEY WORDS:**

**INDEX (***plural***INDICES)****INDEX FORM****BASE**