Exponents clarification question

[hackedagainanda]

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.

 

[FactChecker]

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: $k^{11+6} = k^{17}$

 

[hackedagainanda]

Sorry if I wasn't clear: why is it $\frac {4k^{17}} {125}$ instead of $\frac {4} {125k^{17}}$

 

[haruspex]

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?

 

[Mark44]

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:
$$\frac 2 {125}k^{17}$$

 

[hackedagainanda]

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 $4k^{17} {1/125}$

 

[hackedagainanda]

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.