How to solve indices: Waec 2020 questions on indices

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Table of Content
  • Introduction to solving of waec questions on indices.  
  • What are indices(waec questions on indices) 
  • How to solve waec questions on indices(waec 2020)
  • Solved waec past questions on indices. 
  • A short quiz on indices(Try them out) 

INTRODUCTION(Waec 2020 questions on indices)  

Welcome, this section of my tutorial will be targeted at waec mathematics questions on indices. How to tackle indices mathematics question set by waec, jamb or other exam bodies will be fully expantiated here. The solving of waec questions pertaining  to indices has always been quite easy and straight forward. Indices questions in waec,  jamb(utme), putme are all the same and can be tackled by understanding some basic principles which I will be revealing as we progress. 

Also, most of the questions i will be using as examples will all be waec past questions. This is to show that waec indices questions are quite easy, you can even tackle them with your eyes closed(Lol) . 

The table of content is carefully arranged so as enable you can understand how to tackle WAEC indices questions.

check my tutorial on how to solve linear equations in one variable here.

check my tutorial on how to tackle problems involving surd here

I also have a tutorial on solve simultaneous equations with the substitution method here

With that being said let’s dive right in but first get your writing materials and a glass of juice if you like.  MATH IS FUN.

What are indices?(waec questions on indices)  

According to me, i will describes indices as a mathematical expression that simply consists on numbers carrying other numbers.  Indices are useful when expressing large numbers mathematically.

For example, let’s write one billion in figures:

One billion in figures: 1,000,000,000

One billion in indices: 10^9 

You can clearly see that indices comes very handy when writing out large numbers in mathematics or other science subjects. 

For any expression a^b

a is called the base. 

b is called the power or index.

Generally, an expression 4^7  means: 4 × 4 × 4 × 4 × 4 × 4 × 4.

Meaning, a^b means multiply the base “a” with the same base “a” for “b” number of times easy!! peasy!! 

simply means that when the root of a particular number 

How to solve waec questions on indices (WAEC 2020) 

How to solve waec questions on indices. Solving waec indices questions requires you not only know but understand the laws of indices and how to apply them.

I will be listing and explaining all the laws of indices so that you will be able to tackle waec indices questions. 

Laws of indices. 

The laws of indices only applies to expressions having the same base.

1. Laws of indices #1: a^0 = 1. For this law,  any number or letter raise to the power of 0 is equivalent to one. 

Mathematically,  2^0 = 1, b^0 = 1, (2.6)^0 = 1, (anything)^0 = 1.

2. Laws of indices #2: a^1 = a . Here, anything raise to the power of one remains unchanged.

Meaning,  10^1 = 10, 9^1 = 9

3 Laws of indices #3: a^m × a^n = a^{m+n} , it should be noted that all laws of indices only works for the same base. 

Example; 2^3 × 2^4;

According to the third law, the above question 2^3 × 2^4 can be simplified to give the expression below;

2^3 × 2^4 = 2^{3+4}

2^3 × 2^4 = 2^{7}

4. Law of indices #4: \dfrac{a^m} {a^n} or a^m ÷ a^n = a^{m-n}  

Example: \dfrac{2^6}{2^4}

According to indices law 4,the above expression can be reduced to 2^{6-4};

Simplifying the numerator, gives;


5. Law of indices #5: (a^m) ^n = a^{m×n}

Example: (3^2)^5

According to law 5, the above expression can be simplified to;


Simplifying further;

3^{10} thats the final answer

6. Laws of indices #6: a^{-m} = \dfrac{1}{a^m}

Example: 2^{-3}

2^{-3} = \dfrac{1}{2^3}

Simplifying the denominator further gives;

\dfrac{1}{8}. Final answer

7. Law of indices #7: a^{\frac{1} {n}} = \sqrt[n] {a} :

Where \sqrt[n] {a} : Means nth root of base a

Example; 27^{\frac{1} {3}} :

Following law 7

27^{\frac{1} {3}} = \sqrt[3] {27} :

\sqrt[3] {27}  means the third root of 27. The third root of 27 is 3, since 3×3×3 gives 27.


\sqrt[3] {27} = 3 

Final answer = 3

8. Law of indices #8: a^{\frac{m} {n}} = (\sqrt[n] {a})^m

Example: 8^{\frac{2} {3}}

Following law 8:

8^{\frac{2} {3}} = (\sqrt[3] {8})^2

First let’s reduce the term in bracket. The term in the bracket means the cube root of 8. Since the cube root of 8 is 2.

Therefore ;

8^{\frac{2} {3}} = (2)^2


8^{\frac{2} {3}} = 4

Final answer = 4

9.Laws of indices #9: \left(\dfrac{a} {b}\right) ^m = \dfrac{a^m} {b^m}

Example: \left(\dfrac{2} {3}\right) ^3

Following law 9:

\left(\dfrac{2} {3}\right) ^3 = \dfrac{2^3} {3^3}

Simplifying further;

\left(\dfrac{2} {3}\right) ^3 = \dfrac{8} {27}

Mathematically: \sqrt{a} = \sqrt{b×c} where b must be a prefect square and c can be any integer

Solved waec past questions on indices(How to solve indices) 

1.  Evaluate: \left(4^{\frac{1}{2}} + 125^{\frac{1}{3}} \right)^2(waec 2018)


First we simplify the terms inside the bracket;

Form law 7#, 4^{\frac{1}{2}} = \sqrt[2]{4}  

The right hand side of the equation can be said as “square root of 4”


Form law 7#, 4^{\frac{1}{2}} = 2 

The same can be done to the second term inside the bracket;

From law 7#

125^{\frac{1}{3}} = \sqrt[3]{125}  

The term at the right hand side of the equation means “the cube root of 125”

Since the cube root of 125 is 5;


125^{\frac{1}{3}} = 5  

The expression: \left(4^{\frac{1}{2}} + 125^{\frac{1}{3}} \right)^2 

can now be reduced to 

\left(2+ 5 \right)^2 

Summing up the terms in the bracket gives;

\left(7 \right)^2 

7 raise to the power of two means

7×7 = 49


Final answer = 49

2. Simplify: \dfrac{3^{n-1} × 27^{n+1}}{81^n}


The first step is express 27 and 81 in their index form, having base 3


27 = 3×3×3 = 3^3

81 = 3×3×3×3 = 3^4

Replacing this 27 and 81 with their index form gives

\dfrac{3^{n-1} × (3^3)^{n+1}}{(3^4)^n}

We can reduce the above expression using law #5 to;

\dfrac{3^{n-1} × 3^{3(n+1)}}{3^{4n}}

Clearing the brackets gives;

\dfrac{3^{n-1} × 3^{3n+3}}{3^{4n}}

Consider the numerator of the above expression;

We can use law #3 to reduce the numerator.

Doing this gives;


Collecting like terms at the index part of the numerator gives;


Simplifying the terms at the index section gives


Therefore, we can perform #law 5(division law) on the expression;

Doing this gives;


Collecting like terms gives


Simplifying further gives;


Final answer = 3^2 = 9

3. If \dfrac{27^x×3^{1-x}}{9^{2x}} = 1, find the value of x.(waec 2015)


To solve for x in the expression above, we need to express 27 and 9 to their index form.

Since 27 = 3^3 and 9 = 3^2

We can replace 27 and 9 with their index form in the initial expression.

Therefore, \dfrac{27^x×3^{1-x}}{9^{2x}} = 1 becomes \dfrac{(3^3)^x×3^{1-x}}{(3^2)^{2x}} = 1

Using indices law #5 to reduce the expression further.

\dfrac{3^{3x}×3^{1-x}}{3^{2(2x)}} = 1

Clearing the denominator bracket;

\dfrac{3^{3x}×3^{1-x}}{3^{4x}} = 1

We can use the multiplication law of indices on the numerator of the fraction.

\dfrac{3^{(3x+(1-x)}}{3^{4x}} = 1

Clearing the bracket;

\dfrac{3^{3x+1-x}}{3^{4x}} = 1

Using the division law of indices on the above expression gives:

3^{3x+1-x-4x} = 1

Collecting like terms

3^{3x-x-4x+1} = 1

Simplifying the above expression further gives;

3^{-2x+1} = 1

From the indices laws one, anything raise to the power of 0 is one

It means the the right hand side of the above expression can also written as 2 raise to the power of 0

Replacing one with 2^0

3^{-2x+1} = 2^0

Since the base at the two sides of the equation are the same, it therefore means that the power(index( are also the same, from this conclusion,we can say that

-2x + 1 =  0

Solving the above linear equation gives

-2x = -1

Dividing through by -2 gives

\dfrac{-2x}{-2} = \dfrac{-1}{-2}


x = \dfrac{1}{2}

4. If 2^n = y, find 2^{(2 + \frac{n}{3})}(waec 2015)


Let’s consider the expression at the right hand side

2^{(2 + \frac{n}{3})}

From the multiplication law of indices, we can rewrite the above expression to give


We can reduce the expression using law #5


We can simplify the above expression to give.


Since in the question, 2^n = y

We can replace 2 raise to the power of n with y

Doing this gives;


Final answer is 4y^{\frac{1}{3}}

5. Simplify: \sqrt{\dfrac{8^2×4^{n+1}}{2^{2n}×16}}


First, we can express 4,8 and 16 to their index form;

Since 8 in it’s index form is 2^3

and 16 in it’s index form is 2^4

Replacing 8 and 16 in the expression gives;


Reducing the expression further using #law5


Clearing the bracket


Using the multiplication law of indices


Clearing the bracket and simplifying gives


Using the division law of indices


Clearing the bracket;


Collecting like terms




Since 2^4=16

We can reduce the expression to;


The square root of 16 is 4

Final answer is 4;

I recommend you check my article on the following
Check my tutorial on Word problems leading to linear Equations


My name is John Philip Obhenimen, i am very passionate about educating young people this is because the future is in our hands and our performance is closely tied to the information we consume as young people

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