MATH EXPONENTS - NUMBER OF SOLVED EXAMPLES AND EXERCISES ON EXPONENTS

Please study  Laws of Exponents before Math Exponents,
if you have not already done so.

It is a prerequisite here. There, we stated the 7 laws of indices
and the two Rules used for solving problems.

We also gave the explanations and proofs of the 7 laws.

We provided a few Solved Examples and Problems for Practice with
Answers to help to apply the 7 Laws and the 2 Rules in solving problems.

Here, we provide many more Solved Examples and Exercises with answers.

Studying the worked out problems will help remember and apply
the 7 Laws of exponents and the 2 Rules.

Practice makes one perfect. This is especially true for
remembering Algebra Formulas (Math Formulas).
So, take the exercises seriously and practice solving the problems.

Set of Solved Examples of Math Exponents :
Application of Laws of Exponents

Solved Example 1 of Math Exponents

Evaluate (-1⁄2)⁵⁄(-1⁄2)⁴ + (-1⁄8)⁄(-1⁄4)

Solution to Example 1 of Math Exponents:

We know a^m⁄aⁿ = a^{m - n}
Applying this Law, we get
(-1⁄2)⁵⁄(-1⁄2)⁴ + (-1⁄8)⁄(-1⁄4)
= (-1⁄2)^{5 - 4} + (-1⁄8) x (4⁄-1)
= (-1⁄2)¹ + 1⁄2
= -1⁄2 + 1⁄2 = 0. Ans.

Solved Example 2 of Math Exponents

What is (-1)⁴⁰¹?

Solution to Example 2 of Math Exponents:

Let A = (-1)⁴⁰¹
A = (-1)⁴⁰¹ = (-1) ^{400 + 1}
Interchanging L.H.S. and R.H.S. in Law 1, we have
a^{m + n} = a^m x aⁿ
Applying this here, we get
A = (-1)⁴⁰⁰ x (-1)¹
= (-1)^{2 x 200} x (-1)¹
Interchanging L.H.S. and R.H.S. in Law 2, we have
a^mn = (a^m)ⁿ
Applying this here, we get
A = {(-1)²}²⁰⁰ x (-1)¹
We know (-1)² = (-1) x (-1) = 1.
and (-1)¹ = -1.
∴ A = 1²⁰⁰ x (-1 )
We know 1power any number is 1.
∴ A = 1 x (-1 ) = -1.
Thus (-1)⁴⁰¹ = -1. Ans.

You might have noted (-1)^{odd number} = -1 and (-1)^{even number} = 1.

Solved Example 3 of Math Exponents

Given that 5^x = 1000, find 5^{x + 2} and 5^{x - 2}

Solution to Example 3 of Math Exponents:

Let A = 5^{x + 2} and B = 5^{x - 2}
Interchanging L.H.S. and R.H.S. in Law 1, we have
a^{m + n} = a^m x aⁿ
Applying this here, we get
A = 5^{x + 2} = 5^x x 5²
By data 5^x = 1000. Using this here, we get
A = 1000 x 5² = 1000 x 25 = 25000.
Thus 5^{x + 2} = 25000. Ans.

Interchanging L.H.S. and R.H.S. in Law 4, we have
a^{m - n} = a^m⁄aⁿ
Applying this here, we get
B = 5^{x - 2} = 5^x⁄5²
By data 5^x = 1000. Using this here, we get
B = 1000⁄5² = 1000⁄25 = 40.
Thus 5^{x - 2} = 40. Ans.

Solved Example 4 of Math Exponents

Find x so that (4x)³ = 4³³

Solution to Example 4 of Math Exponents:

Here the exponents are 3 on the L.H.S. and
3³ on the R.H.S.
To find x which is in the base of L.H.S.,
let us make the exponents same.
We know 3³ = 3 x 3 x 3 = 3 x 9 = 9 x 3.
R.H.S. = 4³³ = 4^{9 x 3}
Interchanging L.H.S. and R.H.S. in Law 2, we have
a^mn = (a^m)ⁿ
Applying this here, we get
R.H.S. = 4^{9 x 3} = (4⁹)³
L.H.S. = (4x)³
So, we have
(4x)³ = (4⁹)³
Since the exponents are equal, the bases should be equal.
∴ (4x) = (4⁹)
⇒ x = 4⁹⁄4 = (4⁹)⁄4¹
We know a^m⁄aⁿ = a^{m - n}
Applying this here, we get
x = 4^{9 - 1} = 4⁸. Ans.

Solved Example 5 of Math Exponents

If p⁄q = (2⁄3)³ + (3⁄2)^-3, what is (p⁄q)^-10 ?

Solution to Example 5 of Math Exponents:

We know (a⁄b)^-n = (b⁄a)ⁿ( See Law 3)
Applying this, (3⁄2)^-3 = (2⁄3)³.
By data, p⁄q = (2⁄3)³ + (3⁄2)^-3.
= (2⁄3)³ + (2⁄3)³
= 2{(2⁄3)³}
We know (a⁄b)^m = a^m⁄b^m (See Law 7)
Applying this, (2⁄3)³ = (2³)⁄(3³) = 8⁄27
∴ p⁄q = 2{(2⁄3)³} = 2{8⁄27} = 16⁄27
⇒ (p⁄q)^-10 = (16⁄27)^-10
We know (a⁄b)^-n = (b⁄a)ⁿ.( See Law 3)
Applying this, (16⁄27)^-10 = (27⁄16)¹⁰.
∴ (p⁄q)^-10 = (16⁄27)^-10 = (27⁄16)¹⁰. Ans.

Exercise of Math Exponents :
Application of Laws of Exponents

1. Simplify (i) (4x³)² x (3x³)² x (3xy)⁴ (ii) a^{x + y - z} x a^{y + z - x} x a^{z + x - y}
2. With what should you multiply 2^a to get 2^{a + 2} and 2^{a - 2}
3. Simplify (x^a⁄x^b)^{a + b} x (x^b⁄x^c)^{b + c} x (x^c⁄x^a)^{c + a}

For the Answers, see at the bottom of the page.

Answers to Exercise of Math Exponents :
Application of Laws of Exponents

1. 1296x¹⁰y⁴
2. 1⁄4
3. 1.