Basic Mathematics, Rules for Multiplication trouble

Question

Doing some self study from the text Basic Mathematics by Serge Lang I ran into an exercise question which I can't seem to wrap my head around. The question is:

Express the following expressions in the form $2^m3^na^rb^s$ ,where $m,n,r,s$ are positive integers.

$8a^2b^3(27a^4)(2^5ab)$

After some research I found that the final answer is expressed as

$2^83^3a^7b^4$

I've attempted to use distribution as a means of solving it but end up stuck and confused. I'm entirely lost as to how that answer is derived.

Answer 1

The key idea here is that multiplication is both commutative and associative, so we may multiply in any order and in any groupings we wish, as well as "break up" products into any groupings and any order we wish.

The expression $8a^2b^3(27a^4)(2^5ab)$, by associativity and commutativity, is just the same expression without the parentheses and with the powers of 2 first, then the powers of 3, then the powers of $a$, and finally the powers of $b$. So $8a^2b^3(27a^4)(2^5ab)=8\cdot2^5\cdot27\cdot a^2a^4ab^3b$.

Using our rules of exponents, we can then condense these products of similar numbers into one big power. So $8\cdot2^5\cdot27\cdot a^2a^4ab^3b = 2^83^3a^7b^4$, which is the answer your research uncovered.

Answer 2

You just apply basic power rules:
$a^na^m = a^{n+m}$ etc ...
Let's apply them now to your example: $$8a^2b^3(27)a^2b^3a^4ab$$ $$8 (27) 2^5 a^2 b^3 a^4 ab$$ $$2^3 2^5 (27) a^7 b^4$$ $$2^8 3^3 a^7 b^4$$

Answer 3

I also have another way to do it. If you distribute the $8 a^2 b^3$ with the $27 a^4$, you will get $216 a^6 b^3$. Then, distribute that value with $2^5 a b$ and you will get $6,912 a^7 b^4$. The number $6,912$ can be written exponentially as $(2^8)(3^3)$. Bring that value down with the $a^7 b^4$, and you get your answer.