Algebra Coursebook, Chapter 3, Question 7

[mira.leon]

Algebra Chapter 3, Question 7, page 47:


The question reads:

Simplify: (4^y+4^y+4^y+4^y)(3^y+3^y+3^y)

A) 4^4y * 3^3y B) 12^y+1 C) 16^y * 9^y D) 12^y E) 4^y * 12^y

Can someone please explain how this question simplifies the exponents? I got very confused in trying to figure out how (4^y+4^y+4^y+4^y)(3^y+3^y+3^y) converts to (4^y+1)(3^y+1) ... I could only figure out that it was equivalent to 4(4^y) * 3(3^y).

Thank you!

[RonPurewal]

[quote="mira.leon"]I got very confused in trying to figure out how (4^y+4^y+4^y+4^y)(3^y+3^y+3^y) converts to (4^(y+1))(3^(y+1)) ... I could only figure out that it was equivalent to 4(4^y) * 3(3^y).

The purple thing is the purple thing, and the blue thing is the blue thing.
(I also added parentheses, because, technically, 4^n+1"”without the parentheses"”would mean the sum of 4^n and 1.)

If that's not clear, then consider one (or both) of the following:

1/
Variables in exponents are annoying. I don't like them either. Try replacing them with numbers.
If you have 4^7 times 4, you might be able to see pretty quickly that the result is 4^8. (You're just taking 4*4*4*4*4*4*4 and multiplying another 4.)
Likewise, 4^10 times 4 is 4^11.
And 4^270796 times 4 is 4^270797.
If you can see these things, then that's why 4^n times 4 is 4^(n + 1).

2/
If you know exponent rules really well, then it's sufficient to realize that 4 is the same as 4^1. Then you can multiply (4^1)(4^n) using exponent rules (= add the exponents together).

I like the first approach better because it actually MEANS something"”with the second approach, I'd forget pretty quickly"”but you should use whichever is easier for you.