# 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.

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## 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 aman = am - 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
am + n = am 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
amn = (am)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 5x = 1000, find 5x + 2 and 5x - 2

Solution to Example 3 of Math Exponents:

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

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

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### 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 = 49 x 3
Interchanging L.H.S. and R.H.S. in Law 2, we have
amn = (am)n
Applying this here, we get
R.H.S. = 49 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 aman = am - n
Applying this here, we get
x = 49 - 1 = 48. Ans.

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### Solved Example 5 of Math Exponents

Solved Example 5 of Math Exponents :

If pq = (2⁄3)3 + (3⁄2)-3, what is (pq)-10 ?

Solution to Example 5 of Math Exponents:

We know (ab)-n = (ba)n( See Law 3)
Applying this, (3⁄2)-3 = (2⁄3)3.
By data, pq = (2⁄3)3 + (3⁄2)-3.
= (2⁄3)3 + (2⁄3)3
= 2{(2⁄3)3}
We know (ab)m = ambm (See Law 7)
Applying this, (2⁄3)3 = (23)⁄(33) = 8⁄27
pq = 2{(2⁄3)3} = 2{8⁄27} = 16⁄27
⇒ (pq)-10 = (16⁄27)-10
We know (ab)-n = (ba)n.( See Law 3)
Applying this, (16⁄27)-10 = (27⁄16)10.
∴ (pq)-10 = (16⁄27)-10 = (27⁄16)10. Ans.

### Exercise of Math Exponents : Application of Laws of Exponents

Problems on Math Exponents :

1. Simplify (i) (4x3)2 x (3x3)2 x (3xy)4 (ii) ax + y - z x ay + z - x x az + x - y
2. With what should you multiply 2a to get 2a + 2 and 2a - 2
3. Simplify (xaxb)a + b x (xbxc)b + c x (xcxa)c + a

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

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

Answers to Problems on Math Exponents :

1. 1296x10y4
2. 1⁄4
3. 1.