Explanation and Proof of Law 1 of Exponents with Adding Exponents

Did you answer the question (2) (iv) of Exercise in
Exponents as 5^{3} x 5^{2} = (5 x 5 x 5) x (5 x 5) ? You are right. 5^{3} means 5 is multiplied 3 times. 5^{2} means 5 is multiplied 2 times. 5^{3} x 5^{2} means 5 is multiplied (3 + 2) times. That means 5^{3} x 5^{2} = 5^{(3 + 2)}.

Similarly, (3⁄5)^{4} x (3⁄5)^{5} = (3⁄5 x 3⁄5 x 3⁄5 x3⁄5) x (3⁄5 x 3⁄5 x 3⁄5 x 3⁄5 x 3⁄5) = (3⁄5)^{9} = (3⁄5)^{(4 + 5)}

Generalising this, we get a^{m} x a^{n} = (a x a x a x ...........m times) x (a x a x a x ...........n times) = a x a x a x ...........(m + n) times = a^{(m + n)} where a is any real number and m and n are positive integers.

This gives us Law 1 of Exponents. with adding exponents.

product of powers of the same base:

a^{m} x a^{n} = a^{m + n} where a is any real number and m and n are positive integers.

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You might have written the first step as (4^{5})^{3} = 4^{5} x 4^{5} x 4^{5}.

Here, you might have expanded each factor of 4^{5} as 4 x 4 x 4 x ...5 times and the final answer as 4 x 4 x 4 x ...15 times.

You are right.

But now we have Law 1, which can be applied after the first step as follows. (4^{5})^{3} = 4^{5} x 4^{5} x 4^{5} = 4^{5 + 5 + 5} = 4^{5 x 3} ( since 5 added 3 times is 5 x 3) = 4^{15}.

Did you notice that (4^{5})^{3} = 4^{5 x 3} ?

Here Repeated Adding Exponents lead to Multiplying them.

Let us see one more example.

{(-3⁄4)^{3}}^{4} = (-3⁄4)^{3} x (-3⁄4)^{3} x (-3⁄4)^{3} x (-3⁄4)^{3} = (-3⁄4)^{3 + 3 + 3 + 3} ( By Applying Law 1, above) = (-3⁄4)^{3 x 4} (since 3 added 4 times is 3 x 4)

Again, did you observe that {(-3⁄4)^{3}}^{4} = (-3⁄4)^{3 x 4} ?

Again, Repeated Adding Exponents lead to Multiplying them.

Generalising this, we get (a^{m})^{n} = a^{m} x a^{m} x a^{m} x .......n times = a^{m + m + m + ......n terms} (By Applying Law 1, above) = a^{m x n} (since m added n times is m x n) = a^{mn} where a is any real number and m and n are positive integers.

This gives us Law 2 of Exponents. with adding exponents.

power of a power:

(a^{m})^{n} = a^{mn} where a is any real number and m and n are positive integers.

You may observe: (a^{m})^{n} = a^{mn} = a^{nm} ( since, mn = nmi.e. commutative law) = (a^{n})^{m}.

thus, corollary of Law 2:

(a^{m})^{n} = (a^{n})^{m}.

Thus Law 2 of Adding Exponents is proved.

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