## Basic concepts and techniques in indices problems

The concept of Indices or powers is used extensively in Arithmetic, Trigonometry, Algebra and other branches Mathematics. An example of indices is $2^4=16$. In longhand text, this can be described as 2 to the power 4 is equal to 16.

### Indices is multiplying a number by itself

Conceptually $2^4$ represents a number that results from 2 multiplied by itself 4 times. In other words, $2^4 = 2\times{2\times{2\times{2}}}=16$.

We can then think of raising a number to powers as a **unary operation** (operation on a single operand).

If the power is an integer $n$, it represents, multiplying the number with itself $n$ times. Thus one way to look at indices is to take it as a **compact form of expressing many same factor multiplications.**

### But indices is more than a compact form of many multiplications

But indices or powers is a separate topic by its own right as it creates new possibilities of non-integer powers such as $\frac{1}{2}$ or $0.373$. Because of this generalization, we are able to move out of number domain to abstract algebraic domain and state such equations as,

$a^xa^y = p$.

The **advantage of this** **general form of expression** is obvious. For **any real values** of the variables, this expression holds good. A specific case of this equation can be,

$2^4\times{2^2} = 64$.

**Terminology:** In the term $a^x$, we call $a$ as the **base** and $x$ as the **power**.

### Three basic concepts in the topic of indices

In the expression $2^4\times{2^2} = 64$, where $64=2^6$, two **basic rules (or concepts) of indices** hold good.

#### First basic indices concept

The power of 2 on the Left Hand Side (LHS) has to be equal to the power of 2 on the Right Hand Side because the base in both cases is same, that is, 2. More genrrally, this concept states,

In an equation involving single terms on two sides each with same base raised to powers, the powers must also be same.

For example, if $a^p = a^q$, $p = q$.

In complex problems involving indices, use of this concept many times produces most efficient and elegant solutions.

#### Second basic indices concept

In an expression involving product of terms, powers are added if the base of product terms are same.

As an example,

$2^4\times{2^2} = 2^{4+2} = 2^6$.

This is expressed more generally as,

$a^x\times{a^y} = a^{x + y}$.

This is believable as, $a^x$ has $x$ number of $a$'s multiplied together and likewise $a^y$ has $y$ number of $a$'s multiplied together. When these two are multiplied to each other, we get $x + y$ number of $a$'s multiplied together. This is nothing but $a^{x+y}$.

### Third basic indices concept

**The third rule of indices** we would need for this discussion is,

$(a^x)^y = a^{xy}$.

If we again consider this in terms of individual component factors of $a$'s, this becomes easily visible as the correct expression of a product of $xy$ number of $a$'s. In general, this concept is stated as,

If in a term with a power the whole term is raised to a second power, effectively the powers are multiplied.

There are more basic concepts in indices, but we need these three only for our discussion here.

We will illustrate now the use of **Base equalization technique** by a problem example first.

### Problem example 1.

If $2^{4x}\times{4^{6x}} = 16^8$, $x=?$,

- 2
- 3
- 4
- 8

** Solution:**

Knowing the **first basic concept**, and * sensing* that all the three term bases are in powers of 2, we deduce that

**if we transform the bases on the two sides of the equation to the same value 2**, the

**powers will be equal.**Consequently we would a get

**a linear equation in a single variable**. Finding the value of the unknown variable will then be one simple step more.

Basically this is the **Base equalization technique** and is generally the best technique for solving math problems on indices.

Applying this Base equalization technique, we get,

$2^{4x}\times{4^{6x}} = 2^{4x}\times{2^{12x}}$.

Here we have transformed $4^{6x}$ to $2^{12x}$ by using the knowledge that $4=2^2$ and **the third indices concept**, $(2^2)^{6x} = 2^{12x}$.

Now we apply the **second basic concept in Indices**, $a^x\times{a^y} = a^{x+y}$ giving,

$2^{4x}\times{4^{6x}} = 2^{4x}\times{2^{12x}} = 2^{16x}$.

So from the given equation we get,

$2^{16x} = 16^8 = 2^{32}$,

Again we have applied the **Base equalization technique** and additionally the **third basic concept in indices**, $(a^x)^y = a^{xy}$.

As the *bases* are equal on both sides of the equation, the *powers* also must be equal. This is **application of the first basic concept of indices.** So,

$16x = 32$

Or, $x = 2$

**Answer:** Option a: 2.

**Key concepts used:** Using base equalization technique the complex LHS is transformed in terms of lowest base 2 -- similar transformation for RHS -- bases being equal, powers will also be equal. Use of Basic concepts of indices sums, $a^x\times{a^y} = a^{x+y}$ and $(a^x)^y = a^{xy}$.

We state now **one form of** the **Base equalization technique** *when applied to solve indices problems* as,

In an equation with both sides having powers of single bases, we transform the bases to make them equal, so that equality of the powers correspondingly changed will generate the key to the solution.

**Aside:***In its second form we take the base as the powers and equalize the powers to establish an equality relation between the bases of indices product terms.*

To emphasize the usefulness of this technique we would take up a second problem example.

### Problem example 2.

If $x^{x\sqrt{x}} = (x\sqrt{x})^x$ then $x$ is equal to,

- $\frac{4}{9}$
- $\frac{2}{3}$
- $\frac{9}{4}$
- $\frac{3}{2}$

**Solution**

In indices sums, we would first explore **Base equalization technique**. The LHS is already in powers of $x$. Let's then bring the RHS in powers of $x$,

$x^{x\sqrt{x}} = (x\sqrt{x})^x = (x^\frac{3}{2})^x=x^\frac{3x}{2}$.

Now equating powers on both sides, we get,

$x\sqrt{x}=\frac{3x}{2}$, or,$\sqrt{x}=\frac{3}{2}$,

or $x=\frac{9}{4}$.

**Answer:** Option c: $\frac{9}{4}$.

**Key concepts used:** Equalization of base to $x$ on both sides of the equation -- equating the powers of $x$.

Base equalization technique on indices effectively uses the first basic concept on indices in solving actual problems. That's why it is termed as a technique.

### Efficient problem solving

If you stop to review the two problems and how these were solved quickly, you would notice that,

- Even though the Base equalization technique was applied apparently with great ease, you needed to decide that this technique will solve the problems quickly.
- When the decision to use this technique is confirmed, you still had the job to identify the base value to which all the bases would have to be transformed. We have used the term 'sensing'. That is the key. You should be able to sense from the pattern of the problem
*how exactly*the*relevant*techniques and concepts are to be applied. This 'sense' comes by actually solving a lot of problems systematically.

Furthermore, we observe that all the three basic concepts were used. It means, you must know these concepts and their uses clearly.

More importantly, it also means,

For solving an academic or for that matter a real life problem, you may need to apply more than one technique and concept.

**End note:** It may seem that Base equalization technique is effective only in indices problem solving. Is it so? Give it a thought.

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