Laws of Exponents for Real Numbers

Do you remember how to simplify the following?

`color{blue}((i) 17^2. 17^5=)`

`color{red}((ii) (5^3)7=)`

`color{orange}((iii) (23^(10))/(23^7)=)`

`color{blue}((iv) 7^3. 9^3=)`

Did you get these answers? They are as follows:

`color{green}((i) 17^2. 17^5=17^7)`

`color{blue}((ii) (5^3)7=5^14)`

`color{red}((iii) (23^(10))/(23^7)=23^3)`

`color{red}((iv) 7^3. 9^3=63^3)`

To get these answers, you would have used the following laws of exponents, which you have learnt in your earlier classes. (Here a, n and m are natural numbers.

Remember, a is called the base and m and n are the exponents.)

`color{navy}((i) a^m. a^n = a^m +n)`

`color{blue}((ii) a^(mn) = a^( mn))`

`color{orange}((iii)(a^m)/(a^n) = a^(m-n), m>n)`

`color{blue}((iv)a^mb^m = (ab)m)`

What is (a)0? Yes, it is 1! So you have learnt that `(a)^0 =` So, using (iii), we can get `1/(g^n)=a^(-n)` We can now extend the laws to negative exponents too.

So, for example :

`color{navy}((i) 17^2. 17^(-5)=17^(-3)= 1/(17^3))`

`color{blue}((ii) (5^2)^(-7)=5^(-14))`

`color{green}((iii)(23^(-10)/(23^7)=23^(-17))`

`color{orange}((iv) (7)^(-3).(9)^(-3) = (63)^(-3))`

Suppose we want to do the following computations:

`color{blue}((1) 2^(2/3).2^(1/3))`

`color{orange}((ii)(3^(1/5))^4)`

`color{blue}((iii)(7^(1/5))/(7^(1/3)))`

`color{red}((iv)13^(1/5).17^(1/5))`

It turns out that we can extend the laws of exponents that we have studied earlier, even when the base is a positive real number and the exponents are rational numbers. (Later you will study that it can further to be extended when the exponents are real numbers.)

But before we state these laws, and to even make sense of these laws, we need to first understand what, for example `4^(3/2)` is. So, we have some work to do In Section 1.4, we defined `sqrta` for a real number a > 0 as follows:

Let `a > 0` be a real number and n a positive integer. Then `nsqrta=b` if `bn = a` and `b > 0.`

In the language of exponents, we define `nsqrta= a^(1/n).` So in particular, `3sqrt2 = 2^(1/3)` There are now two ways to look at `4^(3/2)`

`color{blue}(4^(3/2)=(4^(1/2))^3=2^3=8)`

`color{orange}(4^(3/2)=(4^3)^(1/2)=(64)^(1/2)=8)`

Therefore, we have the following definition:

Let a > 0 be a real number. Let m and n be integers such that m and n have no common factors other than 1, and n > 0. Then,

`a^(m/n) = (nsqrta^m) = nsqrt(a^m)`

We now have the following extended laws of exponents:

Let `a > 0` be a real number and p and q be rational numbers. Then, we have

`color{navy}((i)a^p . a^q = a^(p+q))`

`color{blue}((ii)(a^q) = a^(p+q))`

`color{green}((iii) (a^p)/(a^q)=a^(p-q))`

`color{blue}((iv) a^pb^p= (ab)^p)`

`color{blue}((i) 17^2. 17^5=)`

`color{red}((ii) (5^3)7=)`

`color{orange}((iii) (23^(10))/(23^7)=)`

`color{blue}((iv) 7^3. 9^3=)`

Did you get these answers? They are as follows:

`color{green}((i) 17^2. 17^5=17^7)`

`color{blue}((ii) (5^3)7=5^14)`

`color{red}((iii) (23^(10))/(23^7)=23^3)`

`color{red}((iv) 7^3. 9^3=63^3)`

To get these answers, you would have used the following laws of exponents, which you have learnt in your earlier classes. (Here a, n and m are natural numbers.

Remember, a is called the base and m and n are the exponents.)

`color{navy}((i) a^m. a^n = a^m +n)`

`color{blue}((ii) a^(mn) = a^( mn))`

`color{orange}((iii)(a^m)/(a^n) = a^(m-n), m>n)`

`color{blue}((iv)a^mb^m = (ab)m)`

What is (a)0? Yes, it is 1! So you have learnt that `(a)^0 =` So, using (iii), we can get `1/(g^n)=a^(-n)` We can now extend the laws to negative exponents too.

So, for example :

`color{navy}((i) 17^2. 17^(-5)=17^(-3)= 1/(17^3))`

`color{blue}((ii) (5^2)^(-7)=5^(-14))`

`color{green}((iii)(23^(-10)/(23^7)=23^(-17))`

`color{orange}((iv) (7)^(-3).(9)^(-3) = (63)^(-3))`

Suppose we want to do the following computations:

`color{blue}((1) 2^(2/3).2^(1/3))`

`color{orange}((ii)(3^(1/5))^4)`

`color{blue}((iii)(7^(1/5))/(7^(1/3)))`

`color{red}((iv)13^(1/5).17^(1/5))`

It turns out that we can extend the laws of exponents that we have studied earlier, even when the base is a positive real number and the exponents are rational numbers. (Later you will study that it can further to be extended when the exponents are real numbers.)

But before we state these laws, and to even make sense of these laws, we need to first understand what, for example `4^(3/2)` is. So, we have some work to do In Section 1.4, we defined `sqrta` for a real number a > 0 as follows:

Let `a > 0` be a real number and n a positive integer. Then `nsqrta=b` if `bn = a` and `b > 0.`

In the language of exponents, we define `nsqrta= a^(1/n).` So in particular, `3sqrt2 = 2^(1/3)` There are now two ways to look at `4^(3/2)`

`color{blue}(4^(3/2)=(4^(1/2))^3=2^3=8)`

`color{orange}(4^(3/2)=(4^3)^(1/2)=(64)^(1/2)=8)`

Therefore, we have the following definition:

Let a > 0 be a real number. Let m and n be integers such that m and n have no common factors other than 1, and n > 0. Then,

`a^(m/n) = (nsqrta^m) = nsqrt(a^m)`

We now have the following extended laws of exponents:

Let `a > 0` be a real number and p and q be rational numbers. Then, we have

`color{navy}((i)a^p . a^q = a^(p+q))`

`color{blue}((ii)(a^q) = a^(p+q))`

`color{green}((iii) (a^p)/(a^q)=a^(p-q))`

`color{blue}((iv) a^pb^p= (ab)^p)`