Quantitative Aptitude - SPLessons

Exponents and Roots

Chapter 57

SPLessons 5 Steps, 3 Clicks
5 Steps - 3 Clicks

Exponents and Roots

shape Introduction

Exponent:
Denoting the repeated multiplication of a number itself is known as an exponent.

Example: \(3^4\) = 3 x 3 x 3 x 3 = 81.

Here, 3 is the base

4 is the exponent

The expression is read as “3 to the fourth power”.


Root:

The root of a number x is another number, which when multiplied by itself a given number of times, equals x.
Sometimes “root” is used as a quick way of saying “square root”.

Example: \(\sqrt2\) means “root 2”.


shape Methods

Exponents and bases
Suppose in the expression \(a^b\), the bottom number ‘a’ is called the ‘base’ and the top number ‘b’ is called the ‘exponent’.

This particular expression can be read as “a to the power of b” or “a to the fifth power”.

The exponent ‘b’ in the expression, tells of taking the base of ‘a’ and multiplied by itself ‘b’ times.


Example 1:

    \(2^5\)

    Here,

    Base = 2

    Exponent = 5

    The exponent ‘5’ in the expression, tells to take the base of ‘2’ and multiplied by itself ‘5’ times.

    \(2^5\) = 2 x 2 x 2 x 2 x 2 = 32


Example 2:

    \((-4)^3\)

    The exponent ‘3’ in the expression, tells of taking the base of ‘(-4)’ and multiplied by itself ‘3’ times.
    \((-4)^3\) = (-4) x (-4) x (-4) = -64


Example 3:

    \((\frac{2}{3})^4\)

    The exponent ‘4’ in the expression, tells of taking the base of ‘(\(\frac{2}{3}\))’ and multiplied by itself ‘4’ times.

    \((\frac{2}{3})^4\) = \((\frac{2}{3}) * (\frac{2}{3}) * (\frac{2}{3}) * (\frac{2}{3})\) = \(\frac{16}{81}\)


Example 4:

    8 x 8 x 8 x 8 x 8 =?

    As 8 is repeated for five times, given product can be written as \(8^5\).


Example 5:

    13 x 13 x 13 = \((13)^3\)


Example 6:

    (-4) x (-4) = \((-4)^2\)


Example 7:

    x * x * x * x * x * x * x = \((x)^7\)


Example 8:

    \(7^2\) can be read as

    7 to the power of 2 (or) 7 to the second power (or) 7 squared.

    \(7^2\) = 7 x 7 = 49


Example 9:

    \(6^3\) can be read as

    6 to the power of three (or) 6 to the third power (or) 6 cubed.

    \(6^3\) = 6 x 6 x 6 = 216


Properties of exponents

1. Base equals 1:

    \(1^6\) = 1 x 1 x 1 x 1 x 1 x 1 = 1

    In general, 1 raised to any power is always equal to 1.


2. Base equals 0:

    \(0^5\) = 0 x 0 x 0 x 0 x 0 = 0

    In general, 0 raised to any nonzero power is equal to 0.


3. Exponent equals 0:

    \(7^0\) = 1

    If any non-zero number raised to the power of 0 is always equal to 1.


Example 1:

    \(7^0\), \(5^0\), \(100^0\) etc.

    The original definition will also be hard to apply for the expressions which consist of negative exponents.


Example 2:

    \((10)^{-2}\)

    Similarly, not applicable for the expressions like \(8^{\frac{2}{3}}\)


4. Exponent equal to 1:

    \(8^1\) = 8

    In general, any number, ‘x’ raised to the power of 1 is always equal to ‘x’.

    \(x^1\) = x


5. No exponent:

    6 = \(6^1\)

    In fact, if a number has no exponent, then the exponent is assumed to be 1.

    x = \(x^1\)


6. Even and Odd Exponents:

Examples:

    \((-1)^{82}\) = 1

    \((-1)^1\) = -1

    When the exponent is odd, the expression evaluates to be -1.

    When the exponent is even, the expression evaluates to be 1.


    \((-1)^1\) = -1

    \((-1)^2\) = (-1) (-1) = 1

    \((-1)^3\) = (-1) (-1) (-1) = -1

    \((-1)^4\) = (-1) (-1) (-1) (-1) = 1

    and so on..


To generalize the above results,

A negative number raised to even power results in a positive outcome.

A negative number raised to odd power results in a negative outcome.


Building on the above results can say that:

1. An odd exponent preserves the sign of the base.

Example 1: \((-2)^5\) = -32

Example 2: \((10)^3\) = 1000

If the base is negative, the result is also a negative as shown in the first example.

If the base is positive, the result is also a positive as shown in the second example.


2. An even exponent always yields a positive result regardless of the sign of the base. (as long as the base is not equal to zero)


Example 1: \((-3)^4\) = 81

Example 2: \(2^8\) = 256


Powers to memorize

It is quite useful to memorize the certain powers in order to save time while solving.

Powers of 2

    \(2^1\) = 2

    \(2^2\) = 4

    \(2^3\) = 8

    \(2^4\) = 16

    \(2^5\) = 32

    \(2^6\) = 64

    \(2^7\) = 128


Powers of 3

    \(3^1\) = 3

    \(3^2\) = 9

    \(3^3\) = 27

    \(3^4\) = 81


Powers of 4

    \(4^1\) = 4

    \(4^2\) = 8

    \(4^3\) = 64


Powers of 5

    \(5^1\) = 5

    \(5^2\) = 25

    \(5^3\) = 125

    \(5^4\) = 625


Powers of 10

    \(10^1\) = 10

    \(10^2\) = 100

    \(10^3\) = 1000

    \(10^4\) = 10000

    \(10^5\) = 100000

etc……


Exponent laws

1. Product law:

    In general, to find the product of “\(x^a\)” and “\(x^b\)” as the two bases are same and add the exponents i.e. \(x^{a + b}\)

    \((x^a) (x^b)\) = \(x^{a + b}\)


Example 1:

    \(8^5 * 8^2\)

    Here, the two bases are equal,

    \(8^5 * 8^2\) = (8 * 8 * 8 * 8 * 8) * (8 * 8)

    = (8 * 8 * 8 * 8 * 8 * 8 * 8)

    = \(8^7\)


Example 2:

    \(2^3 * 2^7\)

    Here, the two bases are equal,

    \(2^3 * 2^7\) = (2 x 2 x 2) x (2 x 2 x 2 x 2 x 2 x 2 x 2)

    = (2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2)

    = \(2^{10}\)


2. Quotient law:

    In general, if “\(x^a\)” is divided by “\(x^b\)” then the division is equal to \(x^{a – b}\).

    \(\frac{x^a}{x^b}\) = \(x^{a – b}\)


Example 1:

    \(\frac{8^5}{8^2}\) = \(\frac{8 * 8 * 8 * 8 * 8}{8 x 8}\)

    = 8 x 8 x 8

    = \(8^3\)


Example 2:

    \(\frac{2^7}{2^3}\) = \(\frac{2 * 2 * 2 * 2 * 2 * 2 * 2}{2 * 2 * 2}\)

    = 2 x 2 x 2 x 2

    = \(2^4\)


3. Power of a power law:

    In general,

    \({x^a}^b\) = \(x^{ab}\)

    Example 1: \((5^7)^2\) = \((5^7) (5^7)\)

    As the bases are equal, add the powers

    = 56(7 + 7)

    = \(5^{14}\)


Example 2:

    \((2^{10})^3\) = \(2^{10} * 2^{10} * 2^{10}\)

    As the bases are equal, add the powers

    = \(2^{10 + 10 + 10}\)

    = \(2^{30}\)


4. Power of a product law:

Example:

    \((x^{10} y^7)^3\) = \((x^{10} y^7) (x^{10} y^7) (x^{10} y^7)\)

    = \((x^{10}) (x^{10}) (x^{10}) (y^7) (y^7) (y^7)\)

    Apply product law

    = \(x^{30} y^21\)


    By generalizing the results above,

    \((x^ay^b)^n\) = \(x^(an) y^(bn)\)


Example:

    \((w^4 x^2 y^3)^5\) = \(w^{20} x^{10} y^{15}\)


6. Power of a quotient law:

Example:

    \(\frac{x^3}{y^{10}}^5\) = \((\frac{x^3}{y^{10}}) (\frac{x^3}{y^{10}}) (\frac{x^3}{y^{10}}) (\frac{x^3}{y^{10}}) (\frac{x^3}{y^{10}})\)

    = \(\frac{x^{15}}{y^{50}}\)


    By generalizing the results above,

    \((\frac{x^a}{y^b})^n\) = \(\frac{x^{an}}{y^{bn}}\)


Example: \((\frac{k^7}{w^3})^{11}\) = \(\frac{k^{77}}{w^{33}}\)


7. Combining Bases law:

Example:

    \((xy)^3\)

    To simplify this, apply product law

    \((xy)^3\) = \(x^3 y^3\)

    Also can be written as

    \(x^3 y^3\) = \((xy)^3\)

    By generalizing the results above,

    \(x^n y^n\) = \((xy)^n\)


Example:

    \(2^3 * 7^3\) = \((2 * 7)^3\) = \((14)^3\)


Example:

    \((\frac{x}{y})^5 \)

    Apply power of a quotient law

    \((\frac{x}{y})^5 \) = \(\frac{x^5}{y^5}\)

    This can also be written as

    \(\frac{x^5}{y^5}\) = \((\frac{x}{y})^5\)


    To generalize the above results,

    \(\frac{x^n}{y^n}\) = \((\frac{x}{y})^n\)


Example:

    \((\frac{(40)^7)}{8^7}\) = \((\frac{40}{8})^7\) = \(5^7\)


Roots

Square roots:

The square root of ‘n’ is a number that, when squared, is equal to ‘n’.

Example:

    \(\sqrt{49}\) = 7, since \(7^2\) = 49 → Right

    \(\sqrt{49}\) = -7 because \((-7)^2\) = 49 → Wrong


In general, \(\sqrt{n}\) = a number (greater than or equal to zero) that, when squared, equals ‘n’.


Example:

    \(\sqrt{(\frac{4}{25})}\) = \(\frac{2}{5}\), since \((\frac{2}{5})^2\) = \(\frac{4}{25}\)


Properties of square roots

A. If 0 < x < 1 then \(\sqrt{x}\) > x

Example:

    \(\sqrt{0.25}\) > 0.25

    Proof: 0.5 > 0.25


B. If x > 1 then \(\sqrt{x}\) < x

Example:

    \(\sqrt{100}\) < 100

    Proof: 10 < 100


Cube roots

The expression \(\sqrt[3]{n}\) is read as “the third root of n” or “the cube root of n”.

Definition: The cube root of ‘n’ is a number that, when cubed equals to ‘n’.

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

\(\sqrt[3]{(-125)}\) = -5, since \((-5)^3\) = -125


In general, rth root n is a number that, when raised to the power of ‘r’, equals ‘n’.

\(\sqrt[4]{81}\) = 3, since \(3^4\) = 81


Odd roots and even roots

Odd roots: third root, fifth root, seventh root, ninth root, ………..

Even roots: square root, fourth root, sixth root, eighth root, ….


Properties of odd roots:

    1. Can find the odd root of a negative number.

    2. The odd root of a negative number will be negative.

    3. The odd root of a positive number will be positive.


Properties of even roots:

    1. Cannot find the even root of a negative number.

    2. The even root of a positive number will be positive.


Example:

    \(\sqrt[4]{16}\) does not equal 2 and -2

    \(\sqrt[4]{16}\) =2

    Roots as Powers

    \(\sqrt{x}\) = \(x^{\frac{1}{2}}[latex]

    Proof: If x ≥ 0, then [latex](\sqrt{x}) (\sqrt{x})\) = x

    \((\sqrt{x}) (\sqrt{x})\) = \(x^1\)

    let \(\sqrt{x}\) = \(x^k\)

    \((x^k) (x^k)\) = \(x^1\)

    Apply product law

    \(x^{2}\) = \(x^1\)

    2k = 1

    k = \(\frac{1}{2}\)

    Hence, \(\sqrt{x}\) = \(x^{\frac{1}{2}}\)


In general, \(\sqrt[n]{x}\) = \(x^{\frac{1}{n}}\)

Example:

    \((36)^{\frac{1}{2}}\) = \(\sqrt{36}\) = 6 since \(6^2\) = 36


Properties of roots

1. Product of roots and roots of products:
\((\sqrt[n]{x})(\sqrt[n]{y})\) = \(\sqrt[n]{xy}\)


i.e. \(\sqrt[n]{xy}\) = \((\sqrt[n]{x})(\sqrt[n]{y})\)


2. Quotients of roots and roots of quotients:

\(\frac{\sqrt[n]{x}}{\sqrt[n]{y}}\) = \(\sqrt[n]{\frac{x}{y}}\)


i.e. \(\sqrt[n]{\frac{x}{y}}\) = \(\frac{\sqrt[n]{x}}{\sqrt[n]{y}}\)


Operations with roots

1. Addition and subtraction:
Examples:

\( 2\sqrt{7} + 9\sqrt{7}\) = \( 11\sqrt{7}\)


\( 11\sqrt[3]{2} – 5\sqrt[3]{2}\) = \(6\sqrt[3]{2}\)


\( 7\sqrt[5]{3} + 5\sqrt[3]{2} – \sqrt[5]{3} \) = \(8\sqrt[5]{3}\)


2. Multiplication:

Multiply the parts outside the root and multiply the parts inside the root.

Examples: 2\(\sqrt{3}\) x 7\(\sqrt{5}\) = 14\(\sqrt{15}\)


3. Division:

Divide the parts outside the root, and divide the parts inside the root.

Examples: \(\frac{21\sqrt{10}}{7\sqrt{2}}\) = 3\(\sqrt{5}\)

shape Samples

1. If 0 < \(10^n\) < 1,000,000, where n is a non-negative integer, what is the greatest value of \((\frac{1}{2})^n\)?

    A. \(\frac{1}{2}\)
    B. 1
    C. 5
    D. 32
    E. 64


Solution:

    The smaller n becomes the greater \( (\frac{1}{2})^n\) becomes. So what is the smallest value?

    May be tempted to say 1, which would give us \(\frac{1}{2}\). But remember n = 0, because
    \(10^n\) = 1. Therefore \( (\frac{1}{2})^0\)= 1

    Answer is option B.


2. \((\frac{5}{4})^{-n}\) < \(16^{-1}\)
What is the least integer value of n?


Solution:

    The best place to start here is by getting rid of the unseemly negative signs and translating the equation as follows:

    \((\frac{4}{5})^{n}\) < \(\frac{1}{16}\)

    A good little trick to learn using 4/5 taken to some power is that \((\frac{4}{5})^{3}\) = \(\frac{64}{125}\), which is slightly – but only slightly – greater than ½. Therefore, can translate \((\frac{4}{5})^{3}\) to \(\frac{1}{2}\).

    \((\frac{1}{2})^4\)= 1/16

    That would make \((\frac{4}{5})^{12}\) a tad larger than \(\frac{1}{16}\). To make it less than \(\frac{1}{16}\) would multiply by the final \(\frac{4}{5}\), giving us n = 13.


3. Evaluate.

    (i) \(3^{–2}\)
    (ii) \((– 4)^{– 2}\)


Solution:

    (i) \(3^{–2}\)

    = \(\frac{1}{3^2}\)

    = \(\frac{1}{9}\)

    (ii) \((– 4)^{– 2}\)

    = \(\frac{1}{(-4)^2}\)

    = \(\frac{1}{16}\)


4. Evaluate

    (i) \(8^{-1} * 4^3 ÷ 2^{-4}\)
    (ii) \(5^{-1} * 2^{-1}* 6^{-1}\)


Solution:

    (i) \(8^{-1} * 4^3 ÷ 2^{-4}\)

    = \(\frac{1}{8}\) * 16 * 64

    = 128

    (ii) \(5^{-1} * 2^{-1} * 6^{-1}\)

    = \(\frac{1}{5} * \frac{1}{2} * \frac{1}{6}\)

    = \(\frac{1}{60}\)


5. Find the value of m for which \(5^m\) ÷ \(5^{-3}\) = \(5^5\)

Solution:

    Given that

    \(5^m\) ÷ \(5^{-3}\) = \(5^5\)

    Apply, \(a^m\) = \(a^{m-n}\). Then

    Here, m – n = 5 and n = -3

    So, m = 5 + (-3) = 2