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

  • Thread starter 462chevelle
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In summary, the problem is that the homework statement has incorrect fractional exponents. The correct expression is 1/k^4/5 * k^-1.
  • #1
462chevelle
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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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  • #2
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.
 
  • #3
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.
 
  • #4
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
 
  • #5
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:
  • #6
those are fraction exponents. I don't know how to use latex
 
  • #7
here this is crude but it is my first attempt
 

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  • #8
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.
 
  • #9
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.
 

1. What is the first step in solving a simple algebra problem?

The first step in solving a simple algebra problem is to identify the variable in the equation and isolate it on one side of the equal sign.

2. How do I know which operation to use in each step?

You should always follow the order of operations, which is PEMDAS (Parentheses, Exponents, Multiplication/Division, Addition/Subtraction). Start by simplifying any parentheses or exponents, then work from left to right through the remaining operations.

3. What should I do if there are fractions in the equation?

To simplify fractions, find a common denominator and convert all fractions to that denominator. Then, add or subtract the numerators and keep the common denominator. For multiplication and division, simply multiply or divide the numerators and denominators separately.

4. Can I check my answer?

Yes, you can always check your answer by plugging it back into the original equation and solving for the variable. If the solution is correct, both sides of the equation should be equal.

5. What if I get stuck on a step?

If you get stuck on a step, try breaking the problem down into smaller parts and solving each part separately. You can also ask for help from a teacher, tutor, or classmate. It's important to keep trying and not give up!

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