Multiplication & Division With Exponents I. Multiplication
We learnt about exponents here
How much is $2^3 \times 2^4$?
Let us expand this, like below:
$2^3 \times 2^4$
$= (2 \times 2 \times 2) \times (2 \times 2 \times 2 \times 2)$
$= 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
The first number in the multiplication $2^3$ had three $2$s in it and the second number $2^4$ has four $2$s in it. When we multiply all of them together we have seven $2$s in all.
$\texttip{\therefore}{therefore} 2^3 \times 2^4 = 2^7$
Let us take any base $b$ and see how this works.
$b^2 \times b^4$
$= (b \times b) \times (b \times b \times b \times b)$
$= b \times b \times b \times b \times b \times b$ (a total of 6 $b$'s)
$= b^6$
So in general, if you are multiplying two bases with different powers, the powers will add up, and we can write:
$\large{\underline{b^m \times b^n = b^{(m+n)}}}$
Remember, this is true only for same bases, if the bases are different you cannot add the powers or multiply the bases.
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II. Division
Let us take a look at division. We will calculate $3^7 \div 3^2$
$3^7 \div 3^2$
$= \dfrac{3^7}{3^2}$
$= \dfrac{3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3}{3 \times 3}$
we cancel out the two $3$s from the denominator with two $3$s in the numerator.
$= \dfrac{3 \times 3 \times 3 \times 3 \times 3 \times \cancel{3} \times \cancel{3}}{\cancel{3} \times \cancel{3}}$
$= \dfrac{3 \times 3 \times 3 \times 3 \times 3}{1}$
$= 3^5$
So, for any base $b$ we can divide two different powers of $b$, let's say $b^m$ divided by $b^n$, by subtracting the power of the denominator from the power of the numerator, that is:
$\large{\underline{b^m \div b^n = \dfrac{b^m}{b^n} = b^{(m-n)}}}$
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Now let us see how to write an expression of the form $(b^m)^n$
We can expand the outer power $n$ as.
$(b^m)^n = b^m \times b^m \times ...\ n\ times$
Now expanding each of the exponent $m$ we get:
$= (b \times b \times ... m\ times) \times (b \times b \times ... m\ times)... n\ times$
Each group in parenthesis contains $m$ number of $b's$ and there are $n$ such groups.
Therefore, there are a total of $m \times n$ number of $b's$
Therefore, we can write:
$\large{\underline{(b^m)^n = b^{mn}}}$