Exponents clarification question

  • #1
hackedagainanda
50
11
Homework Statement:
Simplify and remove all negative exponents from this equation:

## \frac {2k^2 k^3} {k^{-1} k^{-5}}(5k^{-2})^{-3}##
Relevant Equations:
Quotient Rule: ##\frac {x^a} {x^b} = x^{a-b}##

Product Rule: ## x^a x^b = x^{a+b}##
So my attempt is this: ##(2k)^2 k^3 = 4k^5## to clear the top numerator then to clear the denominator ## k^{-1} k^{-5} = k^{-6}##Then I apply the quotient rule and get ##4k^{11} (5k^{-2})^{-3}## and simplifying the right hand side I get ##5^{-3} k^6## here is where I got lost, why is it that when you use a negative exponent on the 5 to get 1/125 why do you apply the ##k^{17}## to numerator and not the denominator ##\frac {4k^{17}} {125}## Why isn't the answer ## \frac {4} {125k^{17}}## ?

I probably forgot the rules from arithmetic most likely, I feel embarrassed to ask but that's the only way you learn.
 

Answers and Replies

  • #2
FactChecker
Science Advisor
Homework Helper
Gold Member
7,593
3,314
The pieces you did look correct, but there is a danger of losing track of some pieces. It would be better to keep everything together at each step like this:
##\frac {2k^2k^3}{k^{-1}k^{-5}}(5k^{-2})^{-3}##
## = \frac {4k^5}{k^{-1}k^{-5}}(5k^{-2})^{-3}##
## = \frac {4k^5}{k^{-6}}(5k^{-2})^{-3}##
## = 4k^{11}(5k^{-2})^{-3}##
## = 4k^{11}(5^{-3}k^6)##

At this point, I don't see why you are asking your question. You use the negative power for the 5 because you have a negative, ##-3##, and you use the positive power for the ##k^{17}## because you got positive exponents: ##k^{11+6} = k^{17}##
 
  • #3
hackedagainanda
50
11
Sorry if I wasn't clear: why is it ## \frac {4k^{17}} {125}## instead of ##\frac {4} {125k^{17}}##
 
  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
39,212
8,524
Sorry if I wasn't clear: why is it ## \frac {4k^{17}} {125}## instead of ##\frac {4} {125k^{17}}##
You successfully got it to ##4k^{11}5^{-3}k^6##.
What did you do next?
 
  • #5
36,708
8,701
Homework Statement:: Simplify and remove all negative exponents from this equation:

## \frac {2k^2 k^3} {k^{-1} k^{-5}}(5k^{-2})^{-3}##
Relevant Equations:: Quotient Rule: ##\frac {x^a} {x^b} = x^{a-b}##

Product Rule: ## x^a x^b = x^{a+b}##

So my attempt is this: ##(2k)^2 k^3 = 4k^5## to clear the top numerator
Your attempt is not correct. From the original expression, the numerator is ##2k^2k^3##, which is ##2 \cdot k^2 \cdot k^3 = 2k^5##. In your attempt, you have ##(2k)^2k^3##. This is different from the original expression.

BTW, what you're given is not an equation -- an equation has a = symbol in it.
then to clear the denominator ## k^{-1} k^{-5} = k^{-6}##Then I apply the quotient rule and get ##4k^{11} (5k^{-2})^{-3}## and simplifying the right hand side I get ##5^{-3} k^6## here is where I got lost, why is it that when you use a negative exponent on the 5 to get 1/125 why do you apply the ##k^{17}## to numerator and not the denominator ##\frac {4k^{17}} {125}## Why isn't the answer ## \frac {4} {125k^{17}}## ?

I probably forgot the rules from arithmetic most likely, I feel embarrassed to ask but that's the only way you learn.
If the expression in the problem statement is the correct one (i.e., with ##2k^2## rather than ##(2k)^2 )##, then the result I get is this:
$$\frac 2 {125}k^{17}$$
 
Last edited:
  • #6
hackedagainanda
50
11
You successfully got it to ##4k^{11}5^{-3}k^6##.
What did you do next?
I see now, since the first k variable is attached to the 4 I group the second instance of the k variable to the first and get ##4k^{17} {1/125}##
 
  • #7
hackedagainanda
50
11
Your attempt is not correct. From the original expression, the numerator is ##2k^2k^3##, which is ##2 \cdot k^2 \cdot k^3 = 2k^5##. In your attempt, you have ##(2k)^2k^3##. This is different from the original expression.

BTW, what you're given is not an equation -- an equation has a = symbol in it.

If the expression in the problem statement is the correct one (i.e., with ##2k^2## rather than ##(2k)^2 )##, then the result I get is this:
$$\frac 2 {125}k^{17}$$
Sorry I forgot to include the parentheses in the first expression, and you are right the problem is an expression.
 

Suggested for: Exponents clarification question

  • Last Post
Replies
2
Views
560
Replies
32
Views
623
Replies
6
Views
386
Replies
12
Views
441
Replies
13
Views
1K
  • Last Post
Replies
7
Views
82
Replies
2
Views
2K
Replies
8
Views
1K
  • Last Post
Replies
4
Views
366
Replies
8
Views
2K
Top