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.

Get The Best Grades With the Least Amount of Effort

Here is a collection of proven tips, tools and techniques to turn you into a super-achiever - even if you've never thought of yourself as a "gifted" student.

The secrets will help you absorb, digest and remember large chunks of information quickly and easily so you get the best grades with the least amount of effort.

If you apply what you read from the above collection, you can achieve best grades without giving up your fun, such as TV, surfing the net, playing video games or going out with friends!

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}⁄a^{n} = 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 a^{n}
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 = 1^{200} 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 a^{n}
Applying this here, we get
A = 5^{x + 2} = 5^{x} x 5^{2}
By data 5^{x} = 1000. Using this here, we get
A = 1000 x 5^{2} = 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}⁄a^{n}
Applying this here, we get
B = 5^{x - 2} = 5^{x}⁄5^{2}
By data 5^{x} = 1000. Using this here, we get
B = 1000⁄5^{2} = 1000⁄25 = 40.
Thus 5^{x - 2} = 40. Ans.

Here the exponents are 3 on the L.H.S. and
3^{3} 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 3^{3} = 3 x 3 x 3 = 3 x 9 = 9 x 3.
R.H.S. = 4^{33} = 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} = (4^{9})^{3}
L.H.S. = (4x)^{3}
So, we have
(4x)^{3} = (4^{9})^{3}
Since the exponents are equal, the bases should be equal.
∴ (4x) = (4^{9})
⇒ x = 4^{9}⁄4 = (4^{9})⁄4^{1}
We know a^{m}⁄a^{n} = a^{m - n}
Applying this here, we get x = 4^{9 - 1} = 4^{8}. Ans.

Research-based personalized Math Help tutoring program : Math Exponents

Here is a resource for Solid Foundation in Math Fundamentals from Middle thru High School. You can check your self by the

Are you spending lot of money for math tutors to your child and still not satisfied with his/her grades ?

Do you feel that more time from the tutor and more personalized Math Help to identify and fix the problems faced by your child will help ?

Here is a fool proof solution I strongly recommend and that too With a minuscule fraction of the amount you spent on tutors with unconditional 100% money back Guarantee, if you are not satisfied.

It is like having an unlimited time from an excellent Tutor.

It is an Internet-based math tutoring software program that identifies exactly where your child needs help and then creates a personal instruction plan tailored to your child’s specific needs.

If your child can use a computer and access the Internet, he or she can use the program. And your child can access the program anytime from any computer with Internet access.

There is an exclusive, Parent Information Page provides YOU with detailed reports of your child’s progress so you can monitor your child’s success and give them encouragement. These Reports include

Time spent using the program

Assessment results

Personalized remediation curriculum designed for your child

Details the areas of weakness where your child needs additional help

Provides the REASONS WHY your child missed a concept

List of modules accessed and amount of time spent in each module

Quiz results

Creates reports that can be printed and used to discuss issues with your child’s teachers

These reports are created and stored in a secure section of the program, available exclusively to you, the parent. The section is accessed by a password that YOU create and use. No unauthorized users can access this information.

Its research-based results have proven that it really works for all students! in improving math skills and a TWO LETTER GRADE INCREASE in math test scores!,if they invest time in using the program.

Proven for More than 10,000 U.S. public school students who increased their math scores.

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} = (2^{3})⁄(3^{3}) = 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

Problems on Math Exponents :

Simplify (i) (4x^{3})^{2} x (3x^{3})^{2} x (3xy)^{4} (ii) a^{x + y - z} x a^{y + z - x} x a^{z + x - y}

With what should you multiply 2^{a} to get 2^{a + 2} and 2^{a - 2}

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.

Progressive Learning of Math : Math Exponents

Recently, I have found a series of math curricula (Both Hard Copy and Digital Copy) developed by a Lady Teacher who taught everyone from Pre-K students to doctoral students and who is a Ph.D. in Mathematics Education.

This series is very different and advantageous over many of the traditional books available. These give students tools that other books do not. Other books just give practice. These teach students “tricks” and new ways to think.

These build a student’s new knowledge of concepts from their existing knowledge. These provide many pages of practice that gradually increases in difficulty and provide constant review.

These also provide teachers and parents with lessons on how to work with the child on the concepts.

The series is low to reasonably priced and include