Need help to solve this

[Twish]

hello, could someone please explain how to do this question? once I understand it I'm sure I can tackle the whole sheet. thanks!

3^2(-2)^1
2^-1(-3)^2

I forget the rule for numbers with the exponent -1 :?

 

[Denis]

hello, could someone please explain how to do this question? once I understand it I'm sure I can tackle the whole sheet. thanks!
3^2(-2)^1
2^-1(-3)^2
I forget the rule for numbers with the exponent -1 :?

Since 3^2 = 9 and -3^2 = 9, then those cancel out and we're left with:
-2^1 / 2^-1
Since -2^1 = -2, then we're left with: -2 / 2^-1
RULE: a / b^-c = a * b^c
So now we're left with: -2 * 2^1

Can you finish it?

 

[Twish]

wait. sorry, I'm lost. I never learned the part where you cancel things out, isn't there the good ol fashioned way of solving the numerator/denominator first?

 

[stapel]

If, by "solving", you mean "simplifying and/or rearranging", then, yes, you can do that first. This will give you a complex fraction in which the numerator and denominator each contain fractions. Follow the usual procedure for dividing by a fraction (flip and multiply), and the simplify as usual.

Either way will work. Use the method with which you're more comfortable.

Eliz.

 

[opticaltempest]

Hello Twish,

I worked through the problem for you on video. The video may take a few minutes to complete if you are on a slow internet connection. I hope it helps...

http://www.zippyvideos.com/3137639532975626/simplifying_exponents_01-01-06/

 

[Gene]

You can make
2^-1 = 1/(2^1)
or
1/(2^-3) = 2^3
That has the effect of moving across the line.
A^1 = A so

3^2(-2)^1    3^2*(-2)*2^1  9*(-4)
 
---------  = ------------ =  ---- 
2^-1(-3)^2      (-3)^2        9

 

[opticaltempest]

Here is a way to rewrite a term with a negative exponent.

\(\displaystyle \L 2^{ - 1} = \frac{1}{{2^1 }}\)

\(\displaystyle \L 3^{ - 4} = \frac{1}{{3^4 }}\)

\(\displaystyle \L (4x + 2)^{ - 2} = \frac{1}{{(4x + 2)^2 }}\)

The numbers you "push to the bottom" of the fraction with negative exponents cannot be terms that are being added or subtracted. For Example,

\(\displaystyle \L
\frac{{2^{ - 1} + 3}}{{1^2 + 5^{ - 2} }} = \frac{{\left( {\frac{1}{{2^1 }} + 3} \right)}}{{\left( {1^2 + \frac{1}{{5^2 }}} \right)}}\)

Doing something like this would be incorrect,

\(\displaystyle \L
\frac{{2^{ - 1} + 3}}{{1^2 + 5^{ - 2} }} \neq \frac{{3 + 5^2 }}{{1^2 + 2^1 }}\)


Hope this helps some...

 

[soroban]

Hello, Twish!

\(\displaystyle \L\frac{3^2\cdot(-2)^1}{2^{-1}\cdot(-3)^2}\)

I'll give you the "casual" treatment for negative exponents . . .

A term with a negative exponent "gets moved".
\(\displaystyle \;\;\)If it is in the numerator, it moves to the denominator.
\(\displaystyle \;\;\)If it is in the denominator, it moves to the numerator.
In both cases, the exponent changes to positive.

So: \(\displaystyle \L\:\frac{3^2\cdot(-2)^1}{\overbrace{2^{-1}}\cdot(-3)^2}\) becomes: \(\displaystyle \L\frac{\underbrace{2^1}\cdot3^2\cdot(-2)^1}{(-3)^2}\)

And now you can finish it: \(\displaystyle \L\:\frac{2\cdot(3\cdot3)(-2)}{(-3)(-3)} \:=\:\frac{-36}{9}\:=\:-4\)

 

[Denis]

wait. sorry, I'm lost. I never learned the part where you cancel things out, isn't there the good ol fashioned way of solving the numerator/denominator first?

Well, it happens that "the good ol fashioned way" is cancelling things out !

If you were given this problem before being taught cancelling out, then you either
have a BAD teacher... or you missed the class where cancelling was taught

 

[Twish]

wow thank you so much everyone for your help! seriously it's awesome. the video also really helped. thanks a ton guys!