Exponent confusion.

**Forkmaster** · March 22nd 2008, 05:33 PM

Um. i was out the day of the lesson for this assignment, and i read the book's lesson, but it doesn't explain everything. Basically, it says
$a^3 • a^2 = a • a • a • a • a = a^5$
so i learn that a^3 • a^2 (in this situation) is basically adding the exponents. no problem.
then it says...
$(3^5)^2 = 3 to the power of 10$
ok, so you multiply parethetical exponents and outside exponents. again, nothing difficult.
then it says..
$(6 • 5)^2 = 6^2 • 5^2 = 36 • 25 = 900$
add the exponent to everything in parenthesis. again, easy.
but then you get to the problems...
$(3b)^3 • b$
um. so like, you have 3b^3, and multiply it by 1b? so it's... still 3b^3? or did i mess up somewhere?
$5^3 • (5a^4)^2$
5^3=125, so it's 125 • ...5a^8? or 5a^6? @_@
$4x • (x • x^3)^2$
buhhh.... 4x • (x^4)^2.. 4x • x^8... 4x^8?

there's more confusing ones, but hopefully i can sort most of them out if i get these figured out...thanks

**Jhevon** · March 22nd 2008, 05:42 PM

Originally Posted by Forkmaster

then it says..
$(6 • 5)^2 = 6^2 • 5^2 = 36 • 25 = 900$

um, that's wrong, unless you have a typo. as in, you meant to write $(6 \cdot 5)^2$

use the command \cdot to put the multiplication dot in... you can't post images in LaTeX and expect them to work.

$(3b)^3 • b$
um. so like, you have 3b^3, and multiply it by 1b? so it's... still 3b^3? or did i mess up somewhere?

do one piece at a time.

so we have $(3b)^3b$

work out the cubed part first, we get:

$(3^3 b^3)b = (27b^3)b$

now multiply the b's, using the first rule you posted

you get: $27b^4$

now try the others

**topsquark** · March 22nd 2008, 05:42 PM

Originally Posted by Forkmaster

Um. i was out the day of the lesson for this assignment, and i read the book's lesson, but it doesn't explain everything. Basically, it says
$a^3 • a^2 = a • a • a • a • a = a^5$
so i learn that a^3 • a^2 (in this situation) is basically adding the exponents. no problem.
then it says...
$(3^5)^2 = 3 to the power of 10$
ok, so you multiply parethetical exponents and outside exponents. again, nothing difficult.
then it says..
$(6 • 5)^2 = 6^2 • 5^2 = 36 • 25 = 900$
add the exponent to everything in parenthesis. again, easy.
but then you get to the problems...
$(3b)^3 • b$
um. so like, you have 3b^3, and multiply it by 1b? so it's... still 3b^3? or did i mess up somewhere?
$5^3 • (5a^4)^2$
5^3=125, so it's 125 • ...5a^8? or 5a^6? @_@
$4x • (x • x^3)^2$
buhhh.... 4x • (x^4)^2.. 4x • x^8... 4x^8?

there's more confusing ones, but hopefully i can sort most of them out if i get these figured out...thanks

These exponent rules are merely a way of counting factors. Count the number of factors of a in the following "laws" in order to get the rules.
$a^m = \begin{array}{c} \underbrace{(a \cdot a \cdot ~ .... ~ \cdot a)} \\ \text{m times} \end{array}$

$a^m \cdot a^n = \begin{array}{c} \underbrace{(a \cdot a \cdot ~ .... ~ \cdot a)} \\ \text{m times} \end{array} \cdot \begin{array}{c} \underbrace{(a \cdot a \cdot ~ .... ~ \cdot a)} \\ \text{n times} \end{array}$ $= \begin{array}{c} \underbrace{(a \cdot a \cdot ~ .... ~ \cdot a)} \\ \text{m + n times} \end{array} = a^{m + n}$

$(a^m)^n = \begin{array}{c} \underbrace{a^m \cdot ~ ... ~ \cdot a^m} \\ \text{n times} \end{array} = \begin{array}{c} \underbrace{(a \cdot a \cdot ~ .... ~ \cdot a)} \\ \text{mn times} \end{array} = a^{mn}$

$(ab)^m = a^mb^m$
(You can sketch that one out yourself.)

So let's take apart one of your problems.
$(3b)^3 \cdot b$

$= (3^3b^3)b = 3^3 \cdot b^3 \cdot b$

$= 27(b^3 \cdot b^1) = 27b^{3 + 1} = 27b^4$

Always do the parenthesis first.

-Dan

**Gusbob** · March 22nd 2008, 05:43 PM

For $(3b)^3 • b$ :

You can split it into $(3^3) (b^3) (b^1)$ Remember that 3 and b are separate terms in parenthesis, you need to cube both of them.

For $5^3 • (5a^4)^2$ :

Again, 5 and $a^4$ are separate terms, so you need to square both of them, giving you:

$(5^3)(5^2)(a^8)$

**Forkmaster** · March 22nd 2008, 05:55 PM

Originally Posted by Jhevon

um, that's wrong, unless you have a typo. as in, you meant to write $(6 \cdot 5)^2$

yes, i meant to do the math dot. im new to the LaTex thing :P.

thanks for the quick replies. i think i got it now, figured out a few on my own..here's hoping that there's no suckerpunch ones that have entirely new content like this book tends to do, lol

**Forkmaster** · March 22nd 2008, 09:09 PM

Found another one that confused me a bit, but only in one place.
$(-ab)(a^2b)^2$
distribute the ^2 in the second parenthesis...
$(-ab)(a^4b^2)$
now multiply them together...
the b becomes b^3, but what of the a?
-a • a^4...
would it be a^3? -a^4? a to the power of -4? i'm confused.

**Jhevon** · March 22nd 2008, 09:14 PM

Originally Posted by Forkmaster

Found another one that confused me a bit, but only in one place.
$(-ab)(a^2b)^2$
distribute the ^2 in the second parenthesis...
$(-ab)(a^4b^2)$
now multiply them together...
the b becomes b^3, but what of the a?
-a • a^4...
would it be a^3? -a^4? a to the power of -4? i'm confused.

you can think of $-a \cdot a^4$ as $-1 \cdot a \cdot a^4$

so now, just multiply the a's as normal, and then apply the minus 1

**Forkmaster** · March 22nd 2008, 09:15 PM

Got it. Thanks.

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