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# Powers and Roots – Formulas, Examples, Quiz | Maths Tutorial

In this maths tutorial, we introduce exponents / powers and roots using formulas, solved examples and practice questions.

### Powers and Roots | Formulas, Solved Examples, Practice Problems

Exponents, also called powers, are a way of expressing a number multiplied by itself by a certain number of times.

When we write a number a, it is actually a1, said as a to the power 1.

a2 = a*a

a3 = a*a*a
:
:
an = a*a*a*a* . . . n times.

### Basic formulas in Powers and Roots

Some basic formulas used to solve questions on exponents are:

• (am)n = (an)m = amn
• am.an = am+n
• a-m = 1/am
• am/an = am-n = 1/an-m
• (ab)n = anbn
• (a/b)n = an/bn
• a0 = 1

22 = 4. 23 = 8. This is what we learn in exponents.

√4 = 2. 3√8 = 2. This is what we learn in roots.

Here, √ is called the square root or of 2nd order.

3√ is called the cubeth root or of 3rd order.

Similarly we can have the root of a number of any order.

n√a is called a surd of order n.

The symbol n√ is called radical sign,

n is called the order of the surd and

Some basic formulae used to solve questions on roots are:

• n√a = a1/n
• n√ab = n√a* n√b
• n√(a/b) = n√a / n√b
• (n√a)n = a

## Solved examples in Powers & Roots

Let us consider some examples:

Problem 1. Simplify (7.5*105) / (25*10-4)

Solution:

(7.5*105) / (25*10-4)

→ (75*104) / (25*10-4)

Cancelling 75 with 3 times 25 and applying the formula of am/an = am-n

→ 3*104-(-4)

→ 3*108

Problem 2. Find x if 32x-1 + 32x+1 = 270.

Solution:

Taking out a term common, we get

→ 32x-1 (1+32)

Observe that here, we applied the formula am+n = am.an in writing 32x+1 as a product of 32x-1 and 32.

→ 32x-1 (10) = 270

→ 32x-1 = 27

→ 32x-1 = 33

→ 2x-1 = 3

→ x = 2.

Problem 3. Simplify [10 [ (216)1/3 + (64)1/3 ]3 ] 3/4

Solution:

[10 [ (63)1/3 + (43)1/3 ]3 ] 3/4

→ [ 10 [6 + 4]3 ]3/4

→ [ 10 (10)3 ]3/4

→ (104)3/4

→ 103 = 1000.

Problem 4. Simplify [40.08 * (20.22)2 ]10 / [160.16 * (24)0.74 * (42)0.1]

Solution:

[40.08 * (20.22)2]10 / [160.16 * (24)0.74 * (42)0.1]

Applying the formula (am)n = (an)m to the underlined part,

→ [40.08 * (22)0.22]10 / [160.16 * (24)0.74 * (42)0.1]

→ [40.08 * 40.22]10 / [160.16 * (24)0.74 * (42)0.1]

Applying the formula am.an = am+n to the numerator,

→ [40.08+0.22]10 / [160.16 * (24)0.74 * (42)0.1]

Simplifying the denominator,

→ [40.3]10 / [(42)0.16 * (42)0.74 * (42)0.1]

Applying the formula am.an = am+n

→ 43 / [(42)0.16+0.74+0.1]

→ 43 / (42)1

Applying the formula am/an = am-n,

= 4.

Problem 5. Simplify √(5+3√2) + [1/√(5+3√2)]

Solution:

Simplification of this kind of expression also means that the denominator should be rationalized. Rationalizing an expression means removing any square roots present.

The term that rationalizes is called the conjugate. In this example, to rationalize 5+3√2, we use 5-3√2. Hence, 5-3√2 is called the conjugate of 5+3√2 and vice versa.

Consider 5+3√2 = x.

[√x + (1/√x)]2 = x+ 1/x + 2*√x*1/√x

→ (5+3√2) + (1/(5+3√2)) + 2

We have 5+3√2 in the denominator. To remove the square root, we will multiply 1/(5+3√2) with (5-3√2) / (5-3√2). Multiplying with this doesn’t alter the value of the term in any way but helps in rationalization of the denominator and simplification of the expression.

→ (5+3√2) + ((5-3√2) / (5+3√2) (5-3√2)) + 2

Applying the (a+b) (a-b) = a2 – b2 formula to the underlined part,

→ (5+3√2) + [(5-3√2) / (52 – (3√2)2] + 2

→ (5+3√2) + [(5-3√2) / (25 – (9*2)] + 2

→ (5+3√2) + [(5-3√2) / 7] + 2

→ [7(5+3√2) + (5-3√2) + 2(7)] / 7

→ [35 + 21√2 + 5 – 3√2 + 14] / 7

→ [54 + 18√2] / 7

As the original expression was squared to eliminate the roots, we need to apply a square root to this expression.

→ √([54 + 18√2] / 7)

Note: Since we knew the result of the expression will be positive, we were able square and then take the square root the expression. If there is any doubt that it could be negative, then we’d refrain from doing it.

Problem 6. If a2+b2+c2 = ab+bc+ca, simplify [xa/xb]a-b * [xb/xc]b-c * [xc/xa]c-a

Solution:

Applying am/an = am-n, we get

→ (xa-b)a-b * (xb-c)b-c * (xc-a)c-a

Applying the formula (a-b)2 = a2+b2-2ab in the exponent,

→ x(a2 + b2 – 2ab) * x(b2 + c2 – 2bc) * x(c2 + a2 – 2ca)

Applying the am.an = am+n

→ x(a2+b2 – 2ab + b2 + c2 – 2bc + c2 + a2 – 2ca)

→ x(2(a2 + b2 + c2 – (ab + bc + ca)))

→ x(2(0))

→ x0 = 1.

Problem 7. Which is greater: 4√3 or 3√4?

Solution:

In order to compare two surds, they have to be similar i.e., they have to be surds of the same order.

4√3 is a surd of 4th order and 3√4 is a surd of 3rd order.

4√3 can be written as 31/4 and 3√4 as 41/3.

It is still not possible to compare. For this we need to take the LCM of the two orders and express them as surds of one order.

LCM of 3 and 4 is 12.

1/4 can be written as (1/4)*(3/3) = 3/12       AND     1/3 can be written as (1/3)*(4/4) = 4/12.

31/4 can be written as 33/12                                          41/3 can be written as 44/12.

33/12 = (33)1/12 = 12√27                                                  44/12 = (44)1/12 = 12√256

Now, the comparison is between 12√27 and 12√256.

Clearly, 12√256 is greater as 256 > 27.

Therefore, 3√4 > 4√3

### Powers & Roots Quiz: Solve the following problems

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### 6 thoughts on “Powers and Roots – Formulas, Examples, Quiz | Maths Tutorial”

1. Can we write√a-b=√a-√b

• No, you cannot as √25-9 is not equal to √25-√9

2. Solve √7(√7(√7(√7)))