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rational exponents expressed as fractions

CSmith

Member
Aug 22, 2012
41
can someone help me

(4a 3/2)(2a1\2)

(3x5/6)(8x2/3)

(27a6)-2/3

the fractions are powers
 
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Jameson

Administrator
Staff member
Jan 26, 2012
4,040
can someone help me


(4a 3/2)(2a1\2)

(3x5/6)(8x2/3)

(27a6)-2/3
Hi CSmith. We'll be glad to help but you need to try to be more precise with your notation. For example, what does (4a 3/2)(2a1\2) mean?

Is it \(\displaystyle \left(4a \cdot \frac{3}{2} \right) \left(2a \cdot \frac{1}{2} \right)\)?
 

CSmith

Member
Aug 22, 2012
41
they are thge exponents the fractions.written on top of 4a etc...
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
can someone help me

(4a 3/2)(2a1\2)

(3x5/6)(8x2/3)

(27a6)-2/3

the fractions are powers
If the fractions are powers then let's look at the first one.

\(\displaystyle 4a^{\frac{3}{2}} \cdot 2a^{\frac{1}{2}}\)

What you need to do is multiply the constants together and then combine the powers somehow. Here's the trick. If you have \(\displaystyle a^{x} \cdot a^{y}=a^{x+y}\). So when you multiply things in this situation you can add the powers if they have the same base.

What do you get when you try that?

Until you learn how to use Latex on our site I suggest using the ^ sign to indicate exponents. :)
 

CSmith

Member
Aug 22, 2012
41
would it6 ,^3+1=6^4.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
would it6 ,^3+1=6^4.
No, the "a" variable disappeared in your answer. It will still be there! :)

The answer is \(\displaystyle 8a^{2}\) because of the constants, 4*2=8, and when you add the powers you get \(\displaystyle \frac{1}{2}+\frac{3}{2}=\frac{4}{2}=2\). So once again you multiply the constants and add the powers when they have the same base. In this case the base was "a".

Does that make sense?

EDIT: Just to make sure the steps are easy to follow, \(\displaystyle 4a^{\frac{3}{2}} \cdot 2a^{\frac{1}{2}} = 8a^{\frac{1}{2}+\frac{3}{2}}=8a^{\frac{4}{2}}=8a^2\)
 
Last edited:

CSmith

Member
Aug 22, 2012
41
ok thanks

if i have (3x5 ^5/6) (8x^2/3)


would it be 11x how would i work out the 5/6 and the 2/3 in this
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
ok thanks

if i have (3x5 ^5/6) (8x^2/3)


would it be 11x how would i work out the 5/6 and the 2/3 in this
I can't figure out your notation. Does (3x5 ^5/6) mean \(\displaystyle 3x^{\frac{5}{6}}\)? Please put exponents in parentheses like this: x^(5/6).

I'm assuming that (8x^2/3) means \(\displaystyle 8x^{\frac{2}{3}}\)
 

CSmith

Member
Aug 22, 2012
41
Yes
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
Ok, so you have \(\displaystyle 3x^{\frac{5}{6}} \cdot 8x^{\frac{2}{3}}\). This is exactly the same situation as the one I've done for you. You first look at the constants (3 and 8) and multiply them together. Then you look at the variable, "x", and the powers. Since they both have the same base then you just add the powers together like I did in the problem I solved for you. What do you get when you try that?
 

CSmith

Member
Aug 22, 2012
41
ok i got -24x and the base is 6 and 3 right
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
ok i got -24x and the base is 6 and 3 right
Why -24? Yes the constants multiply together to get 8*3=24.

If you have an exponent in the form of \(\displaystyle x^{a}\), x is the base and a is the exponent. When you multiply two things with the same base, x in this case, you add the exponents together. For example \(\displaystyle x^2 \cdot x^5 = x^7\) because 2+5=7.

Your two exponents are \(\displaystyle \frac{5}{6}\) and \(\displaystyle \frac{2}{3}\). You must combine them together by adding.
 

CSmith

Member
Aug 22, 2012
41
yeah i accidently placed the negative.
 

CSmith

Member
Aug 22, 2012
41
24x7/9
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
Closer. \(\displaystyle \frac{5}{6}+\frac{2}{3} \ne \frac{7}{9}\).

\(\displaystyle \frac{2}{3}=\frac{4}{6}\) and now you can add them because they have the same bottom number, or denominator.

\(\displaystyle \frac{5}{6}+\frac{2}{3}=\frac{5}{6}+\frac{4}{6}= \frac{9}{6}=\frac{3}{2}\)

So the final answer is \(\displaystyle 24x^{\frac{3}{2}}\)

Has your teacher gone over how to solve these problems in your class? It looks like you haven't been given sufficient material to know how to do these so it's understandable that you are having trouble. I recommend brushing up on adding and multiplying fractions as well. They can be tough! :)
 

CSmith

Member
Aug 22, 2012
41
well my math teacher only gives out sums to try .Different sums that has several different formats.i feel as if he should focus on one type of situation when dealing with rationals and then focus on the next so that it would be more organized and more easier to catch on.he gives different sums with completly diferent concepts. I am just trying to do the sums in my text book for practice to understand the different ways myself.


how did u get the 2/3 and
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
Here is a video from Khan Academy on adding and subtracting fractions. I recommend watching it to review this material.

The basics of the kinds of problems from your original post are:

1) Multiply constants
2) If two exponents have the same base, then add exponents together
3) Simplify if possible and you have your answer
 

CSmith

Member
Aug 22, 2012
41
Ok thank you sir. i appreciate it
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
Ok thank you sir. i appreciate it
Sure :) If you have other questions please post them. I just think that for this particular kind of question you would do better to watch some videos or read through your textbook and then in a bit coming back to ask any more questions you have on this type of problem.
 

CSmith

Member
Aug 22, 2012
41
khan university really cleared up alot of things