Laws of Exponents dealing with fractions

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• Mar 22nd 2014, 04:02 PM
Joe222
Laws of Exponents dealing with fractions
I have a problem to solve but I am stuck trying to find out how to determine the correct answer. If it was the same variable I would be able to do it but since it's two variables (m and n) I'm not sure how to proceed. I am supposed to re-write it as a single exponent. Please forgive me if the equation format is not using the proper symbols.

Code:

(m/n)^5 (m/n)^-6 (m/n)^0 -------(Fractional Line)-------         (m/n)^20
Thanks

• Mar 22nd 2014, 04:04 PM
fourierT
Re: Laws of Exponents dealing with fractions
An equation implies it is equal to something?
• Mar 22nd 2014, 04:08 PM
Joe222
Re: Laws of Exponents dealing with fractions
Quote:

Originally Posted by fourierT
An equation implies it is equal to something?

Updated the first post post to clarify
• Mar 22nd 2014, 04:14 PM
fourierT
Re: Laws of Exponents dealing with fractions
The mistake is calling it an equation, that is simply a fraction to solve, but anyway.

The powers can be treated the same as if it was one variable seeing as every term in the brackets is the same.

• Mar 22nd 2014, 04:31 PM
Joe222
Re: Laws of Exponents dealing with fractions
Code:

(m/n)^5 (m/n)^-6 (m/n)^0 -------(Fractional Line)-------         (m/n)^20
=

Code:

(m^5/n^5) (m^-6/n^-6) (m/n) -------(Fractional Line)-------- (m^20/n^20)
And this is where I get stuck...

If it were for example:

Code:

(m^5/m^5)
I would know to just subtract the powers but since the variables are different, I don't know the proper way to procede.
• Mar 22nd 2014, 04:37 PM
fourierT
Re: Laws of Exponents dealing with fractions
First of all, you don't need to take the power inside the brackets. Treat (m/n)^5 as you would say x^5 for the first few steps, it's only later that you may need to bring it inside (even then you don't need to but it will be easier for you to see)

Also anything to the power of zero is 1.

Anything to the power of one is itself
• Mar 22nd 2014, 04:56 PM
Joe222
Re: Laws of Exponents dealing with fractions
So something like this then...

Code:

(m/n)^5 (m/n)^-6 (m/n)^0 -------(Fractional Line)------- (m/n)^20
=

Code:

(m/n)^5 (m/n)^-6 (1) -------(Fractional Line)------- (m/n)^20

=

Code:

(m/n 5 - (-6)) (1) -------(Fractional Line)------- (m/n)^20
=

Code:

(m/n)^11 -------(Fractional Line)------- (m/n)^20
=

Code:

(m/n)^11 - 20

=

Code:

(m/n)^-9

Not sure if I'm on the right track here...
• Mar 22nd 2014, 05:06 PM
fourierT
Re: Laws of Exponents dealing with fractions
You are on the right track but you made a mistake, when you multiply terms the exponents are added not taken away.

(m/n)^5 * (m/n)^-6 gives you 5-6 not 5--6 so you get

(m/n)^-1

The next part is correct, you are dividing by (m/n)^20 so you would subtract the 20, correcting your previous error you end up with (m/n)^-21

Do you know how to proceed from here?
• Mar 22nd 2014, 05:13 PM
Joe222
Re: Laws of Exponents dealing with fractions
I think the next step would be to invert from (m/n) to (n/m) so the exponent would become positive?
• Mar 22nd 2014, 05:17 PM
fourierT
Re: Laws of Exponents dealing with fractions
Do you know the steps, or did you just guess that because you have the answer?
• Mar 22nd 2014, 05:19 PM
sakonpure6
Re: Laws of Exponents dealing with fractions
This is as easy as 1,2,3 if you are familiar with these exponential laws.
• 1- When multiplying terms with the same base, we add the exponents
• 2- When dividing terms with the same base, we subtract exponents

Any way,

$\frac{ (\frac{m}{n})^5 \times (\frac{m}{n})^-^6 \times (\frac{m}{n})^0}{ (\frac{m}{n})^2^0 }$

$=\frac{ (\frac{m}{n})^5^-^6^+^0}{\frac{m}{n}^2^1}$

$=(\frac{m}{n})^-^1^-^2^0$

$= (\frac{m}{n}) ^-^2^1$
• Mar 22nd 2014, 05:25 PM
fourierT
Re: Laws of Exponents dealing with fractions
.
• Mar 22nd 2014, 05:25 PM
fourierT
Re: Laws of Exponents dealing with fractions
Quote:

Originally Posted by sakonpure6
This is as easy as 1,2,3 if you are familiar with these exponential laws.
• 1- When multiplying terms with the same base, we add the exponents
• 2- When dividing terms with the same base, we subtract exponents

Any way,

$\frac{ (\frac{n}{m})^5 \times (\frac{n}{m})^-^6 \times (\frac{n}{m})^0}{ (\frac{n}{m})^2^0 }$

$=\frac{ (\frac{n}{m})^5^-^6^+^0}{\frac{n}{m}^2^1}$

$=(\frac{n}{m})^-^1^-^2^0$

$= (\frac{n}{m}) ^-^2^1$

This is wrong, the terms are (m/n) not (n/m).
• Mar 22nd 2014, 05:30 PM
Joe222
Re: Laws of Exponents dealing with fractions
Quote:

Originally Posted by fourierT
Do you know the steps, or did you just guess that because you have the answer?

Well, it's a logical guess on my part because that's what I recall regarding negative exponents as a final answer need to be changed into positive exponents.