# Dividing exponents, Why don't I use BEDMAS for this?

1. Jun 25, 2012

### Gregory.gags

Okay, I'm doing an advanced functions course online to get into Physics for university. I got this question in the exponents section...

(2x3y7)5/(2x2y5)7

Intuition (BEDMAS) tells me to divide out whats IN the brackets (2/2 , x3/x2 , y7/y5) , and then the exponents outside the brackets (5/7 I know this doesn't make that much sense now that I think back to it...) So I looked at the answer and what I should have done, but what I don't understand is WHY i am supposed to START with the exponents (on the outside) INSTEAD of dividing whats inside the brackets, doesn't that go against BEDMAS? This may sound quite silly to someone who has a natural understanding of numbers but i can't see the logic behind it ( other than that you can't divide 5 by 7 :P )

2. Jun 25, 2012

### Dick

What is BEDMAS supposed to mean and how does it supercede the other laws of algebra?

3. Jun 25, 2012

### DrewD

The order of operations tells you to do brackets/parentheses first. But the parens are not combined together. They are two separate expressions and can't be further simplified. In order to further simplify you need to get rid of the parens. Since the exponents distribute, you can try that. In this case it gets rid of the first step in BEDMAS (or PEMDAS where I'm from) and next group like terms and simplify the exponents.

What you actually tried to do was skip past B and E and straight to D. When you divided 2/2 you had to ignore the Brackets/Parens and the Exponents in order to do the Division. My guess is that it seemed right because if felt like you were evaluating all of the Brackets/Parens together. That seems logical, but the actual operation that you were doing was division.

I hope I'm right about what you were thinking and that helps.

4. Jun 26, 2012

### HallsofIvy

Staff Emeritus
Are you clear on what inside the brackets means? 2/2 , x3/x etc. are not inside any brackets. The two brackets (more correctly "parentheses") are $(2x^3y^7)$ and $(2x^7y^2)$.

Doing the brackets first means doing $(2x^3y^7)^5= 32x^{15}y^{35}$ and $$(2x^7y^5)^7= 128x^{49}y^{35}$$. You are NOT using BEDMAS when you Divide before taking the Exponential of the Brackets.

"BEDMAS" is mnemonic for the order of operations: Brackets,Exponents,Division,Multiplication,Addition,Subtraction.

I believe that is commonly used in Britain. In the United States we are typically taught "Please Excuse My Dear Aunt Sally" with "P" for "parentheses" and "M" and "D" reversed (of course they can be done in either order).

Last edited: Jun 26, 2012
5. Jun 26, 2012

### skiller

I'm 42 from England, and was always taught BODMAS. I only heard of BEDMAS a short time ago, so maybe the teaching system has changed in the last 20 years. (Presumably the powers that be thought that "Order" would confuse children when the whole thing is about order, so they changed it to "Exponent".)

I like your American "Please Excuse My Dear Aunt Sally". Not heard that before.

6. Jun 26, 2012

### HallsofIvy

Staff Emeritus
Actually I started to write "BODMAS", then noticed that Gregory.gags had written "BEDMAS".

7. Jun 26, 2012

### Gregory.gags

Thanks guys, you've all been a great help! I was just mistaken but I did it over again (the right way) and got the right answer, so thanks again :)

8. Jun 28, 2012

### Staff: Mentor

I really don't know what I was taught. :uhh: But as far back as I can recall I've assumed the "O" stood for "of", as in "one-third of ...". So your reference to "Order" meant nothing to me (What is order?). But a quick google search reassured me that I'm not alone with the "of" interpretation; this author considers it "of", too. http://www.key2study.com/bodmas.htm

9. Jun 28, 2012

### skiller

Not sure how "of" as in "one-third of" makes much sense to me, in this context. Isn't "one-third of" just a multiplication (or division depending on how you look at it)?

"Order" is just another word relating to "exponent" or "index" or "power". So, for example in the expression a(b+c)2, you would square (b+c) first before multiplying by 2.
This page you've referenced doesn't mention powers so I'm not convinced of its validity really.