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

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 a^{1}, said as a to the power 1.

a^{2} = a*a

a^{3} = a*a*a

:

:

a^{n} = a*a*a*a* . . . n times.

Some basic formulas used to solve questions on exponents are:

- (a
^{m})^{n}= (a^{n})^{m}= a^{mn} - a
^{m}.a^{n }= a^{m+n} - a
^{-m }= 1/a^{m} - a
^{m}/a^{n}= a^{m-n}= 1/a^{n-m} - (ab)
^{n}= a^{n}b^{n} - (a/b)
^{n}= a^{n}/b^{n} - a
^{0}= 1

2^{2} = 4. 2^{3} = 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 2^{nd} order.

^{3}√ is called the cubeth root or of 3^{rd} 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

a is called the radicand.

Some basic formulae used to solve questions on roots are:

^{n}√a = a^{1/n}^{n}√ab =^{n}√a*^{ n}√b^{n}√(a/b) =^{n}√a /^{n}√b- (
^{n}√a)^{n}= a

Let us consider some examples:

**Problem 1.** Simplify (7.5*10^{5}) / (25*10^{-4})

__Solution__:

(7.5*10^{5}) / (25*10^{-4})

→ (75*10^{4}) / (25*10^{-4})

Cancelling 75 with 3 times 25 and applying the formula of a^{m}/a^{n} = a^{m-n}

→ 3*10^{4-(-4)}

→ 3*10^{8}

**Problem 2.** Find x if 3^{2x-1 }+ 3^{2x+1} = 270.

__Solution__:

Taking out a term common, we get

→ 3^{2x-1 }(1+3^{2})

Observe that here, we applied the formula a^{m+n} = a^{m}.a^{n} in writing 3^{2x+1} as a product of 3^{2x-1} and 3^{2}.

→ 3^{2x-1} (10) = 270

→ 3^{2x-1} = 27

→ 3^{2x-1 }= 3^{3}

→ 2x-1 = 3

→ x = 2.

**Problem 3.** Simplify [10 [ (216)^{1/3 }+ (64)^{1/3} ]^{3} ] ^{3/4}

__Solution__:

[10 [ (6^{3})^{1/3 }+ (4^{3})^{1/3} ]^{3} ] ^{3/4}

→ [ 10 [6 + 4]^{3} ]^{3/4}

→ [ 10 (10)^{3} ]^{3/4}

→ (10^{4})^{3/4}

→ 10^{3} = 1000.

**Problem 4.** Simplify [4^{0.08} * (2^{0.22})^{2} ]^{10} / [16^{0.16} * (2^{4})^{0.74 }* (4^{2})^{0.1}]

__Solution__:

[4^{0.08} * __(2 ^{0.22})^{2}__]

Applying the formula (a^{m})^{n} = (a^{n})^{m } to the underlined part,

→ [4^{0.08} * (2^{2})^{0.22}]^{10} / [16^{0.16} * (2^{4})^{0.74 }* (4^{2})^{0.1}]

→ [4^{0.08 }* 4^{0.22}]^{10 }/ [16^{0.16} * (2^{4})^{0.74 }* (4^{2})^{0.1}]

Applying the formula a^{m}.a^{n} = a^{m+n }to the numerator,

→ [4^{0.08+0.22}]^{10} / [16^{0.16} * (2^{4})^{0.74 }* (4^{2})^{0.1}]

Simplifying the denominator,

→ [4^{0.3}]^{10 } / [(4^{2})^{0.16} * (4^{2})^{0.74} * (4^{2})^{0.1}]

Applying the formula a^{m}.a^{n} = a^{m+n}

→ 4^{3 }/ [(4^{2})^{0.16+0.74+0.1}]

→ 4^{3 }/ (4^{2})^{1}

Applying the formula a^{m}/a^{n} = a^{m-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) = a^{2 }– b^{2 }formula to the underlined part,

→ (5+3√2) + [(5-3√2) / (5^{2} – (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 a^{2}+b^{2}+c^{2} = ab+bc+ca, simplify [x^{a}/x^{b}]^{a-b} * [x^{b}/x^{c}]^{b-c} * [x^{c}/x^{a}]^{c-a}

__Solution__:

Applying a^{m}/a^{n }= a^{m-n}, we get

→ (x^{a-b})^{a-b} * (x^{b-c})^{b-c} * (x^{c-a})^{c-a}

Applying the formula (a-b)^{2} = a^{2}+b^{2}-2ab in the exponent,

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

Applying the a^{m}.a^{n} = a^{m+n}

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

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

→ x^{(2(0))}

→ x^{0 }= 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 4^{th} order and ^{3}√4 is a surd of 3^{rd} order.

^{4}√3 can be written as 3^{1/4 }and ^{3}√4 as 4^{1/3}.

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.

3^{1/4} can be written as 3^{3/12} 4^{1/3} can be written as 4^{4/12}.

3^{3/12} = (3^{3})^{1/12} = ^{12}√27 4^{4/12} = (4^{4})^{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__:

Problem 1

If (2^{10}.2^{n}.4^{3}) / (8^{n}.16) = 16, find n.

A. 3

B. 2

C. 5

D. 4

Answer 1

D.

**Explanation**

(2^{10}.2^{n}.(2^{2})^{3}) / (2^{3})^{n}.2^{4} = 2^{4}

Applying the formula a^{m}/a^{n} = a^{m-n }and taking only the exponents,

→ 16+n – (3n+4) = 4

→ 12+n – 3n – 4 = 4

Problem 2

If x^{a + b + c} = 3, find the value of (x^{2a – b} / x^{-b}) . (x^{2b – c} / x^{-c}) . (x^{2c – a} / x^{-a})

A. 3

B. 6

C. 9

D. None

Answer 2

C.

**Explanation**:

Applying a^{m}/a^{n} = a^{m-n},

→ x^{2a – b + b} . x^{2b – c + c} . x^{2c – a + a}

Applying a^{m}.a^{n }= a^{m+n},

→ x^{2a+2b+2c}

→ (x^{a+b+c})^{2}

→ 3^{2 }= 9

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## 2 Comments

Can we write√a-b=√a-√b

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