- Thread starter
- #1

- Thread starter CSmith
- Start date

- Thread starter
- #1

- Admin
- #2

- Jan 26, 2012

- 4,040

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?can someone help me

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

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

(27a6)-2/3

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

- Thread starter
- #3

- Admin
- #4

- Jan 26, 2012

- 4,040

If the fractions are powers then let's look at the first one.can someone help me

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

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

(27a6)-2/3

the fractions are powers

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

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.

- Thread starter
- #5

- Admin
- #6

- Jan 26, 2012

- 4,040

No, the "a" variable disappeared in your answer. It will still be there!would it6 ,^3+1=6^4.

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:

- Thread starter
- #7

- Admin
- #8

- Jan 26, 2012

- 4,040

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

- Thread starter
- #9

- Admin
- #10

- Jan 26, 2012

- 4,040

- Thread starter
- #11

- Admin
- #12

- Jan 26, 2012

- 4,040

Why -24? Yes the constants multiply together to get 8*3=24.ok i got -24x and the base is 6 and 3 right

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.

- Thread starter
- #13

- Thread starter
- #14

- Admin
- #15

- Jan 26, 2012

- 4,040

Closer. \(\displaystyle \frac{5}{6}+\frac{2}{3} \ne \frac{7}{9}\).24x7/9

\(\displaystyle \frac{2}{3}=\frac{4}{6}\) 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!

- Thread starter
- #16

how did u get the 2/3 and

- Admin
- #17

- Jan 26, 2012

- 4,040

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

- Thread starter
- #18

- Admin
- #19

- Jan 26, 2012

- 4,040

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.Ok thank you sir. i appreciate it

- Thread starter
- #20