Solving a Simple Algebra Problem: Step-by-Step Guide

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The discussion revolves around simplifying the expression (k^(1/5))/k^(4/5) * k^-1. Participants clarify that the expression can be simplified to k^(-8/5), confirming the initial calculation. There is confusion regarding the notation used, particularly with the division format, leading to different interpretations of the problem. Suggestions are made to simplify the denominator first and to correctly apply exponent rules. Ultimately, the consensus is that the original approach was correct, and the issue stemmed from misinterpretation of the notation.
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Homework Statement



(k^(1/5))/k^4/5 * k^-1

Homework Equations





The Attempt at a Solution

step 1(k^(1/5))/k^4/5 * k^-1
step 2: (k^1/5)/(k^4/5) * (k^5/5) changed from neg exp. 1/k^1 is implied there
step 3: (k^1/5)/(k^9/5)
answer: 1/(k^8/5)
I have to be making a stupid mistake. but this is such a simple problem.. can anyone help me figure out where I am messing this up?
thanks
 
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462chevelle said:

Homework Statement



(k^(1/5))/k^4/5 * k^-1

Homework Equations





The Attempt at a Solution

step 1(k^(1/5))/k^4/5 * k^-1
step 2: (k^1/5)/(k^4/5) * (k^5/5) changed from neg exp. 1/k^1 is implied there
step 3: (k^1/5)/(k^9/5)
answer: 1/(k^8/5)
I have to be making a stupid mistake. but this is such a simple problem.. can anyone help me figure out where I am messing this up?
thanks

Looks OK to me if I'm reading your problem correctly, which I assume is this:
$$ \frac{k^{1/5}}{k^{4/5}} k^{-1}$$

This simplifies to k-8/5, which is equal to what you have.
 
I may have typed that wrong but off the top of my head I don't see where it would matter. but the k^-1 is in the denominator after k^4/5. theyre multiplied.
 
462chevelle said:

Homework Statement



(k^(1/5))/k^4/5 * k^-1
I have no idea what this is supposed to mean. a/b/c is NOT standard mathematics notation. If I try to interpret the "/" as division it could still be interpreted as (a/b)/c or as a/(b/c) which are NOT the same thing.

If you mean (k^(1/5)/k^4)/5k^-1 then k^(1/5)/k^4= k^(1/5- 4)= k^(1/5- 20/5)= k^(-19/5). Then (k^(1/5)/k^4)/5k^-1= (k^(-19/5)/5k^-1= (1/5)k^(-19/5+1)= (1/5)k^(-14/5).

But k^4/5k^-1= (1/5)k^(4+1)= (1/5)k^5. So k^1/5/(k^4/5k^-1)= (k^1/5)/((1/5)k^5= 5 k^(1/5- 5)= 5k^(1/5- 25/5)= 5k^(-24/5)

Homework Equations





The Attempt at a Solution

step 1(k^(1/5))/k^4/5 * k^-1
step 2: (k^1/5)/(k^4/5) * (k^5/5) changed from neg exp. 1/k^1 is implied there
step 3: (k^1/5)/(k^9/5)
answer: 1/(k^8/5)
I have to be making a stupid mistake. but this is such a simple problem.. can anyone help me figure out where I am messing this up?
thanks
 
disregard this. I donno what i was thinking>>>>>>>wait. I just redid it like that and it is 1/k^4/5 I think
second step would be
(k^1/5)/(k^4/5)(1/k)
third would be (k^1/5)/((k^4/5)/(k^5/5))
is that correct for that format?
 
Last edited:
those are fraction exponents. I don't know how to use latex
 
here this is crude but it is my first attempt
 

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Apparently this is the problem:
$$\frac{k^{1/5}}{k^{4/5}k^{-1}} $$

Simplify the denominator first. What do you get? Note that k-1 ≠ k5/5.
 
here is another attempt. though I don't think its correct. the first attempt seems correct to me.
 

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  • #10
462chevelle said:
here is another attempt. though I don't think its correct. the first attempt seems correct to me.
That's correct, but you're taking the long way in getting to it.

Instead of writing k-1 as 1/k in the denominator, simply add the exponents to get k-1/5 in the denominator. The idea is that ar * as = ar + s.
 
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  • #11
does it not equal 1/k^5/5 though? because it would equal 1/k^1?? I was just using 5/5 for the lcd of the denominator but ill try again.
 
  • #12
so this is how youre saying I should do it?
 

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  • #13
put it in and that was correct. thanks! knew it was a stupid mistake.
 
  • #14
462chevelle said:
here this is crude but it is my first attempt
You appear to have turned ##k^{-1}## into ##k^{\frac 55}##.
LaTex is pretty easy for something like this. Type a double hash symbol (two times '#') at each end of the expression. To raise to a power use ^, as you did in the OP, but if the exponent consists of more than one item, as here, you need to wrap it in curly braces, {}. Fractions take the form \frac{ }{ }, with the numerator inside the first curly braces and the denominator inside the second pair. Again, you can omit a pair of braces if there's only one item inside. Click the Quote button to see how I wrote it above. Always click Preview Post before submitting a Post with LaTex.
One gotcha - if you notice a mistake after posting and click Edit, when you submit ("save") the edited form it will not interpret the LaTex on your screen. The post is probably ok, but it will look wrong to you. Just do a screen refresh.
 
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