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Laws of Exponents for Real Numbers
  • Last Updated : 01 Apr, 2021

Sometimes we come across very large numbers like the mass of earth. Mass of the earth is 5,970,000,000,000, 000, 000, 000, 000 kg. There is no problem in writing mass of earth like this, but sometimes during calculations involving such objects. It becomes cumbersome and hard to use it. That’s why exponents were invented. To make it easy to deal with very big or very small numbers. This number can be written as 5.97 × 1024 Kg. It is read 5.97 times 10 raised to the power 24. Let’s study some laws regarding these exponent numbers. 

Laws of Exponents

These laws help us to simplify our calculations. 

Product law

The Product law states that if the base of two numbers is same, their exponents can be added directly.

 am × an = am+n

To verify this property, see how many times “a” is multiplied in total. It is multiplied “m + n” times. Thus, the property. 



Example: a2a3 = (aa)(aaa) = a5

Quotient law

According to this law, if the two numbers in the numerator and denominator are the same, their exponents can be arranged in such a manner that the exponent in the denominator is subtracted by the exponent in the numerator 

 \frac{a^m}{a^n} = a^{m-n}

This can also be verified similar to the previous one, just see the number of times a is multiplied and then reduce that number by the number of times it is divided. 

Example: \frac{x^4}{x^2}  =\frac{xxxx}{xx} = x^{4-2} = x^2 

Power Law

According to Power-law, if an exponent is a power of another exponent, we can simply multiply the exponents.

(am)n = amn

First, multiply “a” m times, and then do this operation n times. 



Example: (x3)2 = (xxx)2 = (xxx)(xxx) = x6 = x(3 × 2)

Power of Product law

According to the power of product law, if two real numbers say, a and b here are multiplied and raised to power m, we can distribute the exponent to both a and b separately.

 am x bm = (ab)m

This property is just a rearrangement of all these variables. 

Example: a3 × b3 = (aaa) (bbb) = (ab)(ab)(ab) = (ab)3

Power of Quotient law

According to the power of quotient law, if two real numbers are in numerator and denominator and are raised to power n, the power can be separately distributed to both the numbers.

 (\frac{x}{y})^n = \frac{x^n}{y^n}

This also can be verified by a simple rearrangement of the variables.

Example: \frac{x^3}{y^3} = \frac{xxx}{yyy} = \frac{x}{y}\frac{x}{y}\frac{x}{y} = (\frac{x}{y})^{3}

Zero Power Rule

As long as x is not equal to zero, raising it to the power of zero should give us 1 as result.



a0 = 1, a≠ 1. 

Let’s take an example to make this more evident, 

1 = 1 = \frac{x^a}{x^a}  = x^{a - a} = x^0

Note:

Expression 00 is considered to be indeterminate. Why? Because there are two answers to it. 

We know, x0 = 1, So, 00 = 1. 

We also know, 0x = 0. So, 00 = 0. 

These are two contradictory answers, thus we consider 00 to be indeterminate. 

Exponent of Exponent

Sometimes in more complex scenarios, exponents over exponents are given. Let’s see how to approach them, 

5^{2^{4}}

We will start solving from up, 

5^{2^{4}} \\ = 5^{16}

Let’s look at some examples and questions where these properties are applicable. 

Sample Problems

Question 1: Find the value of 2-3

Answer:

  2^{-3}\\ = \frac{1}{2^3}\\ = \frac{1}{8}

Question 2: \frac{1}{3^{-2}}

Answer: 

\frac{1}{3^{-2}} \\ = 3^{-2} \\ = \frac{1}{3^2}  \\= \frac{1}{9}

Question 3: Simplify using above properties: 

(-4)5 x (-4)-10

Answer:

(-4)5 x (-4)-10

= -4(5-10)                     (am × an = am+n)

= (-4)-5

\\ = \frac{1}{(-4)^5}\\ \hspace{0.6cm} (a^{-m} = \frac{1}{a^m}) \\ = \frac{1}{-1024}

Question 4: Simplify using above properties: 

2^5 \div 2^{-6}

Answer: 

2^5 \div 2^{-6} \hspace{0.6cm} (\frac{a^m}{a^n} = a^{m-n})\\ = \frac{2^5}{2^{-6}} \\ = 2^{5 + (-6)} \\ = 2^{-1} \\ = \frac{1}{2}

Question 5: Simplify, (x^3 \div x^{\frac{1}{2}}) \times (x^{\frac{3}{2}} \div x^{0}) \times x^7

Answer:

 (x^3 \div x^{\frac{1}{2}}) \times (x^{\frac{3}{2}} \div x^{0}) \times x^7 \\ = \frac{x^3}{x^{\frac{1}{2}}} \times \frac{x^{\frac{3}{2}}}{x^0} \times x^7 \\ = x^{3 - \frac{1}{2}} \times x^{\frac{3}{2} - 0} \times x^{7} \\ = x^{\frac{5}{2}} \times x^{\frac{3}{2}} \times x^{7} \\ = x^{\frac{5}{2} + \frac{3}{2} + 7}  \\ = x^{\frac{5 + 3+ 14}{2}}  \\ = x^{\frac{22}{2}}\\ = x^{11}

Question 6: What is the 5th root of 515

Answer: 

We know that nth of a number “a” is represented by a^{\frac{1}{n}}

Thus, 5th of 515 will be given by, 

5^{\frac{15}{5}}

= 53

= 125

Question 7: Find the value of n 

2^{n-3} = \frac{1}{16^{n-4}}

Answer: 

2^{n-3} = \frac{1}{16^{n-4}} \\ 2^{n-3} = \frac{1}{2^{4(n-4)}}

2n-3 × 24n – 16 = 1 

2n – 3 + 4n – 16 = 20

Comparing Both Sides

n – 3 + 4n -16 = 0 

5n -19 = 0 

\\ n = \frac{19}{5}

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