 POSCHOLARS How to solve indices; Waec 2023 questions on indices

Oct 26 2022 1:31:00 PM

John Philip

# WAEC MATHEMATICS QUESTIONS ON INDICES

1. Introduction to solving Waec questions on indices
2. What are indices(Waec questions on indices)
3. How to solve Waec questions on indices(Waec 2023)
4. Solved Waec past questions on indices(How to solve indices)
5. A short quiz on indices(Try them out)
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## Introduction to solving Waec questions on indices

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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.
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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).
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The table of content is carefully arranged so as enable you can understand how to tackle WAEC indices questions.
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check my tutorial on how to solve linear equations in one variable here.
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check my tutorial on how to tackle problems involving surd here
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I also have a tutorial on solving simultaneous equations with the substitution method here
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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.
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## 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.
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For example, let's write one billion in figures:
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One billion in figures: 1,000,000,000
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One billion in indices: $10^9$
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You can clearly see that indices come in very handy when writing out large numbers in mathematics or other science subjects.
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For any expression $a^b$
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a is called the base.
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b is called the power or index.
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Generally, an expression $4^7$ means: 4 × 4 × 4 × 4 × 4 × 4 × 4.
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Meaning, $a^b$ means multiply the base "a" with the same base "a" for "b" a number of times easily!! peasy!!
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simply means that when the root of a particular number
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## How to solve Waec questions on indices(Waec 2023)

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

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The laws of indices only apply to expressions having the same base.
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1. Laws of indices #1: $a^0 = 1$. For this law,  any number or letter raised to the power of 0 is equivalent to one.
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Mathematically,  $2^0 = 1, b^0 = 1, (2.6)^0 = 1, (anything)^0 = 1$.
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2. Laws of indices #2: $a^1 = a$. Here, anything raised to the power of one remains unchanged.
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Meaning,  $10^1 = 10, 9^1 = 9$
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3. 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.
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Example; $2^3 × 2^4$;
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According to the third law, the above question $2^3 × 2^4$ can be simplified to give the expression below;
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$2^3 × 2^4 = 2^{3+4}$
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$2^3 × 2^4 = 2^{7}$
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4. Law of indices #4: $\dfrac{a^m} {a^n}$ or $a^m ÷ a^n = a^{m-n}$
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Example: $\dfrac{2^6}{2^4}$
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According to indices law 4, the above expression can be reduced to $2^{6-4}$;
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Simplifying the numerator gives;
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$2^2$
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5. Law of indices #5: $(a^m) ^n = a^{m×n}$
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Example: $(3^2)^5$
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According to law 5, the above expression can be simplified to;
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$3^{2×5}$
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Simplifying further;
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$3^{10}$ that's the final answer
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6. Laws of indices #6: $a^{-m} = \dfrac{1}{a^m}$
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Example: $2^{-3}$
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$2^{-3} = \dfrac{1}{2^3}$
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Simplifying the denominator further gives;
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$\dfrac{1}{8}$. Final answer
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7. Law of indices #7: $a^{\frac{1} {n}} = \sqrt[n] {a}$ :
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Where $\sqrt[n] {a}$ : Means nth root of base a
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Example; $27^{\frac{1} {3}}$ :
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Following law 7
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$27^{\frac{1} {3}} = \sqrt {27}$ :
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$\sqrt {27}$  means the third root of 27. The third root of 27 is 3 since 3×3×3 gives 27.
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Therefore;
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$\sqrt {27} = 3$
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8. Law of indices #8: $a^{\frac{m} {n}} = (\sqrt[n] {a})^m$
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Example: $8^{\frac{2} {3}}$
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Following law 8:
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$8^{\frac{2} {3}} = (\sqrt {8})^2$
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First, 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.
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Therefore ;
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$8^{\frac{2} {3}} = (2)^2$
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Finally;
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$8^{\frac{2} {3}} = 4$
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9. Laws of indices #9: $\left(\dfrac{a} {b}\right) ^m = \dfrac{a^m} {b^m}$
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Example: $\left(\dfrac{2} {3}\right) ^3$
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Following law 9:
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$\left(\dfrac{2} {3}\right) ^3 = \dfrac{2^3} {3^3}$
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Simplifying further;
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$\left(\dfrac{2} {3}\right) ^3 = \dfrac{8} {27}$
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Mathematically: $\sqrt{a} = \sqrt{b×c}$ where b must be a perfect square and c can be any integer
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## Solved Waec past questions on indices(How to solve indices)

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1.  Evaluate: $\left(4^{\frac{1}{2}} + 125^{\frac{1}{3}} \right)^2$(Waec 2018)
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Solution
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First, we simplify the terms inside the bracket;
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Form law 7#, $4^{\frac{1}{2}} = \sqrt{4}$
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The right-hand side of the equation can be said as "square root of 4"
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Therefore;
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Form law 7#, $4^{\frac{1}{2}} = 2$
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The same can be done to the second term inside the bracket;
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From law 7#
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$125^{\frac{1}{3}} = \sqrt{125}$
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The term at the right-hand side of the equation means "the cube root of 125"
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Since the cube root of 125 is 5;
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Therefore;
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$125^{\frac{1}{3}} = 5$
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The expression:$\left(4^{\frac{1}{2}} + 125^{\frac{1}{3}} \right)^2$
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can now be reduced to
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$\left(2+ 5 \right)^2$
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Summing up the terms in the bracket gives;
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$\left(7 \right)^2$
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7 raised to the power of two means
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7×7 = 49
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Therefore,
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2. Simplify: $\dfrac{3^{n-1} × 27^{n+1}}{81^n}$
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Solution
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The first step is express 27 and 81 in their index form, having base 3
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Therefore;
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$27 = 3×3×3 = 3^3$
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$81 = 3×3×3×3 = 3^4$
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Replacing 27 and 81 with their index form gives
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$\dfrac{3^{n-1} × (3^3)^{n+1}}{(3^4)^n}$
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We can reduce the above expression using law #5 to;
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$\dfrac{3^{n-1} × 3^{3(n+1)}}{3^{4n}}$
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Clearing the brackets gives;
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$\dfrac{3^{n-1} × 3^{3n+3}}{3^{4n}}$
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Consider the numerator of the above expression;
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We can use law #3 to reduce the numerator.
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Doing this gives;
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$\dfrac{3^{(n-1)+3n+3}}{3^{4n}}$
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Collecting like terms at the index part of the numerator gives;
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$\dfrac{3^{n+3n+3-1}}{3^{4n}}$
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Simplifying the terms in the index section gives
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$\dfrac{3^{4n+2}}{3^{4n}}$
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Therefore, we can perform #law 5(division law) on the expression;
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Doing this gives;
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$3^{4n+2-4n}$
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Collecting like terms gives
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$3^{4n-4n+2}$
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Simplifying further gives;
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$3^2$
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Final answer = $3^2 = 9$
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3. If $\dfrac{27^x×3^{1-x}}{9^{2x}} = 1$, find the value of x.(Waec 2015)
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Solution:
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To solve for x in the expression above, we need to express 27 and 9 in their index form.
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Since $27 = 3^3 and 9 = 3^2$
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We can replace 27 and 9 with their index form in the initial expression.
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Therefore, $\dfrac{27^x×3^{1-x}}{9^{2x}} = 1$ becomes $\dfrac{(3^3)^x×3^{1-x}}{(3^2)^{2x}} = 1$
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Using indices law #5 to reduce the expression further.
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$\dfrac{3^{3x}×3^{1-x}}{3^{2(2x)}} = 1$
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Clearing the denominator bracket;
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$\dfrac{3^{3x}×3^{1-x}}{3^{4x}} = 1$
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We can use the multiplication law of indices on the numerator of the fraction.
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$\dfrac{3^{(3x+(1-x)}}{3^{4x}} = 1$
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Clearing the bracket;
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$\dfrac{3^{3x+1-x}}{3^{4x}} = 1$
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Using the division law of indices on the above expression gives:
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$3^{3x+1-x-4x} = 1$
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Collecting like terms
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$3^{3x-x-4x+1} = 1$
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Simplifying the above expression further gives;
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$3^{-2x+1} = 1$
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From the indices laws one, anything raised to the power of 0 is one
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It means the right-hand side of the above expression can also be written as 2 raised to the power of 0
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Replacing one with $2^0$
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$3^{-2x+1} = 2^0$
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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
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-2x + 1 =  0
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Solving the above linear equation gives
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-2x = -1
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Dividing through by -2 gives
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$\dfrac{-2x}{-2} = \dfrac{-1}{-2}$
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Therefore,
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$x = \dfrac{1}{2}$
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4. If $2^n = y$, find $2^{(2 + \frac{n}{3})}$(Waec 2015)
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Solution:
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Let's consider the expression on the right-hand side
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$2^{(2 + \frac{n}{3})}$
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From the multiplication law of indices, we can rewrite the above expression to give
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$2^2×2^{\frac{n}{3}}$
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We can reduce the expression using law #5
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$2^2×(2^n)^{\frac{1}{3}}$
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We can simplify the above expression to give.
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$4(2^n)^{\frac{1}{3}}$
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Since in the question, $2^n = y$
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We can replace 2 raised to the power of n with y
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Doing this gives;
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$4(y)^{\frac{1}{3}}$
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Final answer is $4y^{\frac{1}{3}}$
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5. Simplify: $\sqrt{\dfrac{8^2×4^{n+1}}{2^{2n}×16}}$
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Solution
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First, we can express 4,8 and 16 to their index form;
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Since 8 in its index form is $2^3$
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and 16 in its index form is $2^4$
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Replacing 8 and 16 in the expression gives;
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$\sqrt{\dfrac{(2^3)^2×(2^2)^{n+1}}{2^{2n}×2^4}}$
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Reducing the expression further using #law5
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$\sqrt{\dfrac{2^{2(3)}×2^{2(n+1)}}{2^{2n}×2^4}}$
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Clearing the bracket
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$\sqrt{\dfrac{2^6×2^{2n+2}}{2^{2n}×2^4}}$
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Using the multiplication law of indices
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$\sqrt{\dfrac{2^{6+(2n+2)}}{2^{2n+4}}}$
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Clearing the bracket and simplifying gives
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$\sqrt{\dfrac{2^{2n+8}}{2^{2n+4}}}$
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Using the division law of indices
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$\sqrt{2^{2n+8-(2n+4)}}$
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Clearing the bracket;
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$\sqrt{2^{2n+8-2n-4}}$
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Collecting like terms
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$\sqrt{2^{2n-2n+8-4}}$
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Simplifying;
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$\sqrt{2^{4}}$
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Since $2^4=16$
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We can reduce the expression to;
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$\sqrt{16}$
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The square root of 16 is 4
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### I recommend you check my article on the following:

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

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This is all we can take on “How to solve indices; Waec 2023 questions on indices“.
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