# WAEC MATHEMATICS QUESTIONS ON INDICES

###### Table of Contents

paragraph

## Introduction to solving Waec questions on indices

paragraph

Welcome, this section of my tutorial will be targeted at Waec mathematics questions on indices. How to tackle indices mathematics questions set by Waec, Jamb or other exam bodies will be fully expatiate here. The solving of Waec questions pertaining to indices has always been quite easy and straightforward. 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.
paragraph

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).
paragraph

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

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

**here.**paragraph

check my tutorial on how to tackle problems involving surd

**here**paragraph

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

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.
paragraph

## What are indices(Waec questions on indices)

I will describe indices as a mathematical expression that simply consists of numbers carrying other numbers. Indices are useful when expressing large numbers mathematically.

paragraph

For example, let's write one billion in figures:
paragraph

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

One billion in indices: $10^9$
paragraph

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

For any expression $a^b$
paragraph

a is called the base.
paragraph

b is called the power or index.
paragraph

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

Meaning, $a^b$ means multiply the base "a" with the same base "a" for "b" a number of times easily!! peasy!!
paragraph

simply means that when the root of a particular number
paragraph

## How to solve Waec questions on indices(Waec 2023)

paragraph

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

paragraph

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

### Laws of indices.

paragraph

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

- Laws of indices #1: $a^0 = 1$. For this law, any number or letter raised to the power of 0 is equivalent to one.
paragraphMathematically, $2^0 = 1, b^0 = 1, (2.6)^0 = 1, (anything)^0 = 1$.paragraph
- Laws of indices #2: $a^1 = a$. Here, anything raised to the power of one remains unchanged.
paragraphMeaning, $10^1 = 10, 9^1 = 9$paragraph
- Laws of indices #3: $a^m × a^n = a^{m+n}$, it should be noted that all laws of indices only work for the same base.
paragraphExample; $2^3 × 2^4$;paragraphAccording to the third law, the above question $2^3 × 2^4$ can be simplified to give the expression below;paragraph$2^3 × 2^4 = 2^{3+4}$paragraph$2^3 × 2^4 = 2^{7}$paragraph
- Law of indices #4: $\dfrac{a^m} {a^n}$ or $a^m ÷ a^n = a^{m-n}$
paragraphExample: $\dfrac{2^6}{2^4}$paragraphAccording to indices law 4, the above expression can be reduced to $2^{6-4}$;paragraphSimplifying the numerator gives;paragraph$2^2$paragraph
- Law of indices #5: $(a^m) ^n = a^{m×n}$
paragraphExample: $(3^2)^5$paragraphAccording to law 5, the above expression can be simplified to;paragraph$3^{2×5}$paragraphSimplifying further;paragraph$3^{10}$ that's the final answerparagraph
- Laws of indices #6: $a^{-m} = \dfrac{1}{a^m}$
paragraphExample: $2^{-3}$paragraph$2^{-3} = \dfrac{1}{2^3}$paragraphSimplifying the denominator further gives;paragraph$\dfrac{1}{8}$. Final answerparagraph
- Law of indices #7: $a^{\frac{1} {n}} = \sqrt[n] {a}$ :
paragraphWhere $\sqrt[n] {a}$ : Means nth root of base aparagraphExample; $27^{\frac{1} {3}}$ :paragraphFollowing law 7paragraph$27^{\frac{1} {3}} = \sqrt[3] {27}$ :paragraph$\sqrt[3] {27}$ means the third root of 27. The third root of 27 is 3 since 3×3×3 gives 27.paragraphTherefore;paragraph$\sqrt[3] {27} = 3$paragraphFinal answer = 3paragraph
- Law of indices #8: $a^{\frac{m} {n}} = (\sqrt[n] {a})^m$
paragraphExample: $8^{\frac{2} {3}}$paragraphFollowing law 8:paragraph$8^{\frac{2} {3}} = (\sqrt[3] {8})^2$paragraphFirst, let's reduce the term to bracket. The term in the bracket means the cube root of 8. Since the cube root of 8 is 2.paragraphTherefore ;paragraph$8^{\frac{2} {3}} = (2)^2$paragraphFinally;paragraph$8^{\frac{2} {3}} = 4$paragraphFinal answer = 4paragraph
- Laws of indices #9: $\left(\dfrac{a} {b}\right) ^m = \dfrac{a^m} {b^m}$
paragraphExample: $\left(\dfrac{2} {3}\right) ^3$paragraphFollowing law 9:paragraph$\left(\dfrac{2} {3}\right) ^3 = \dfrac{2^3} {3^3}$paragraphSimplifying further;paragraph$\left(\dfrac{2} {3}\right) ^3 = \dfrac{8} {27}$.paragraphMathematically: $\sqrt{a} = \sqrt{b×c}$ where b must be a perfect square and c can be any integerparagraph

## Solved Waec past questions on indices(How to solve indices)

paragraph

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

**Solution**paragraph

First, we simplify the terms inside the bracket;
paragraph

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

The right-hand side of the equation can be said as "square root of 4"
paragraph

Therefore;
paragraph

Form law 7#, $4^{\frac{1}{2}} = 2$
paragraph

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

From law 7#
paragraph

$125^{\frac{1}{3}} = \sqrt[3]{125}$
paragraph

The term at the right-hand side of the equation means "the cube root of 125"
paragraph

Since the cube root of 125 is 5;
paragraph

Therefore;
paragraph

$125^{\frac{1}{3}} = 5$
paragraph

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

can now be reduced to
paragraph

$\left(2+ 5 \right)^2$
paragraph

Summing up the terms in the bracket gives;
paragraph

$\left(7 \right)^2$
paragraph

7 raised to the power of two means
paragraph

7×7 = 49
paragraph

Therefore,
paragraph

Final answer = 49
paragraph

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

**Solution**paragraph

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

Therefore;
paragraph

$27 = 3×3×3 = 3^3$
paragraph

$81 = 3×3×3×3 = 3^4$
paragraph

Replacing 27 and 81 with their index form gives
paragraph

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

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

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

Clearing the brackets gives;
paragraph

$\dfrac{3^{n-1} × 3^{3n+3}}{3^{4n}}$
paragraph

Consider the numerator of the above expression;
paragraph

We can use law #3 to reduce the numerator.
paragraph

Doing this gives;
paragraph

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

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

$\dfrac{3^{n+3n+3-1}}{3^{4n}}$
paragraph

Simplifying the terms in the index section gives
paragraph

$\dfrac{3^{4n+2}}{3^{4n}}$
paragraph

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

Doing this gives;
paragraph

$3^{4n+2-4n}$
paragraph

Collecting like terms gives
paragraph

$3^{4n-4n+2}$
paragraph

Simplifying further gives;
paragraph

$3^2$
paragraph

Final answer = $3^2 = 9$
paragraph

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

**Solution:**paragraph

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

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

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

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

Using indices law #5 to reduce the expression further.
paragraph

$\dfrac{3^{3x}×3^{1-x}}{3^{2(2x)}} = 1$
paragraph

Clearing the denominator bracket;
paragraph

$\dfrac{3^{3x}×3^{1-x}}{3^{4x}} = 1$
paragraph

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

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

Clearing the bracket;
paragraph

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

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

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

Collecting like terms
paragraph

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

Simplifying the above expression further gives;
paragraph

$3^{-2x+1} = 1$
paragraph

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

It means the right-hand side of the above expression can also be written as 2 raised to the power of 0
paragraph

Replacing one with $2^0$
paragraph

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

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
paragraph

-2x + 1 = 0
paragraph

Solving the above linear equation gives
paragraph

-2x = -1
paragraph

Dividing through by -2 gives
paragraph

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

Therefore,
paragraph

$x = \dfrac{1}{2}$
paragraph

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

**Solution:**paragraph

Let's consider the expression on the right-hand side
paragraph

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

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

$2^2×2^{\frac{n}{3}}$
paragraph

We can reduce the expression using law #5
paragraph

$2^2×(2^n)^{\frac{1}{3}}$
paragraph

We can simplify the above expression to give.
paragraph

$4(2^n)^{\frac{1}{3}}$
paragraph

Since in the question, $2^n = y$
paragraph

We can replace 2 raised to the power of n with y
paragraph

Doing this gives;
paragraph

$4(y)^{\frac{1}{3}}$
paragraph

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

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

**Solution**paragraph

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

Since 8 in its index form is $2^3$
paragraph

and 16 in its index form is $2^4$
paragraph

Replacing 8 and 16 in the expression gives;
paragraph

$\sqrt{\dfrac{(2^3)^2×(2^2)^{n+1}}{2^{2n}×2^4}}$
paragraph

Reducing the expression further using #law5
paragraph

$\sqrt{\dfrac{2^{2(3)}×2^{2(n+1)}}{2^{2n}×2^4}}$
paragraph

Clearing the bracket
paragraph

$\sqrt{\dfrac{2^6×2^{2n+2}}{2^{2n}×2^4}}$
paragraph

Using the multiplication law of indices
paragraph

$\sqrt{\dfrac{2^{6+(2n+2)}}{2^{2n+4}}}$
paragraph

Clearing the bracket and simplifying gives
paragraph

$\sqrt{\dfrac{2^{2n+8}}{2^{2n+4}}}$
paragraph

Using the division law of indices
paragraph

$\sqrt{2^{2n+8-(2n+4)}}$
paragraph

Clearing the bracket;
paragraph

$\sqrt{2^{2n+8-2n-4}}$
paragraph

Collecting like terms
paragraph

$\sqrt{2^{2n-2n+8-4}}$
paragraph

Simplifying;
paragraph

$\sqrt{2^{4}}$
paragraph

Since $2^4=16$
paragraph

We can reduce the expression to;
paragraph

$\sqrt{16}$
paragraph

The square root of 16 is 4
paragraph

The final answer is 4;
paragraph

### I recommend you check my article on the following:

**Check my tutorial on surd****How to pass chemistry in the wassce 2020 exams with an A1****How to blast mathematics in the wassce 2020 exams with an A****How to get an A in the wassce biology 2020 exams****How to blast physics in the wassce 2023 exams with an A.****How to pass the wassce further mathematics exam excellently well in 2020****How to manage time efficiently as a student.****Seven mistakes to avoid as a year one students in the university.**paragraph

#### Check my tutorial on Word problems leading to linear Equations

paragraph

paragraph

This is all we can take on

**“How to solve indices; Waec 2023 questions on indices“.**paragraph