something is not right here: Simplify [(8 x^4 y^{-2})/(4 x^7)]^{-3}

[allegansveritatem]

I spent an hour on this today and could not get it to come out the way the book says it should. kI turned it every way I could think of and reviewed the past chapter on negative exponents and still something is awry. What?

I am asked to simplify this:


And this is what I finally came to and what I should have come to:

 

[Deleted member 4993]

I spent an hour on this today and could not get it to come out the way the book says it should. kI turned it every way I could think of and reviewed the past chapter on negative exponents and still something is awry. What?

I am asked to simplify this:
View attachment 10096

And this is what I finally came to and what I should have come to:

View attachment 10097

2^(-3) = 1/8

 

[topsquark]

I spent an hour on this today and could not get it to come out the way the book says it should. kI turned it every way I could think of and reviewed the past chapter on negative exponents and still something is awry. What?

I am asked to simplify this:
View attachment 10096

And this is what I finally came to and what I should have come to:

View attachment 10097

\(\displaystyle (2x^{-3})^{-3} = (2)^{-3}(x^{-3}) = \dfrac{x^9}{8}\)

You forgot the - sign in \(\displaystyle 2^{-3}\)

-Dan

Whoops! Subhotosh Khan beat me to it.

 

[Denis]

I am asked to simplify this:
View attachment 10096

Start by simplifying what's inside the brackets:
8x^4 y^(-2) / 4x^7 = 2 / (x^3 y^2)
So yer now at [2 / (x^3 y^2)]^(-3)
Continue...

AND don't forget to assign values to x and y (like x=2 and y=3)
so you can check your work after each step...

 

[mmm4444bot]

Start by simplifying what's inside the brackets …

allegansveritatem's work shows they did that. The mistake also shown: negative exponent doesn't m͏ean negative power.

 

[sinx]

I spent an hour on this today and could not get it to come out the way the book says it should. kI turned it every way I could think of and reviewed the past chapter on negative exponents and still something is awry. What?

I am asked to simplify this:
View attachment 10096

And this is what I finally came to and what I should have come to:

View attachment 10097

others have caught your mistake, i.e. 2^-3=1/8
it is sometimes easier to get rid of negative exponents, e.g.
(2/x³y²)^-3 =(x³y²/2)³ = x⁹y⁶/8

 

[mmm4444bot]

… sometimes easier to get rid of negative exponents [first] …

This thread shows how three steps (that is, thought process) may be completed in different order.

\(\displaystyle \bigg( \dfrac{2y^{-2}}{x^3} \bigg)^{-3}\\
\;\\
\bigg( \dfrac{\frac{1}{8} y^6}{x^{-9}} \bigg)\\
\;\\
\bigg( \dfrac{x^9 y^6}{8} \bigg)\)

 

[allegansveritatem]

2^(-3) = 1/8

I see it! I did a number on that two when I should have left it with its exponent, inverted it and changed the sign.
Thanks for the tip.

 

[allegansveritatem]

Start by simplifying what's inside the brackets:
8x^4 y^(-2) / 4x^7 = 2 / (x^3 y^2)
So yer now at [2 / (x^3 y^2)]^(-3)
Continue...

AND don't forget to assign values to x and y (like x=2 and y=3)
so you can check your work after each step...

Well, actually I did a lot of assigning values to the variables in order to figure out what I could do with the factors when dividing.

 

[allegansveritatem]

This thread shows how three steps (that is, thought process) may be completed in different order.

\(\displaystyle \bigg( \dfrac{2y^{-2}}{x^3} \bigg)^{-3}\\
\;\\
\bigg( \dfrac{\frac{1}{8} y^6}{x^{-9}} \bigg)\\
\;\\
\bigg( \dfrac{x^9 y^6}{8} \bigg)\)

so, whenever there is a fraction in the numerator do you invert it and put it in the denominator?

 

[allegansveritatem]

so, whenever there is a fraction in the numerator do you invert it and put it in the denominator?

thanks to all who replied. Fractions with exponents throw me a little. I have to go back and do the problems again after the chapter that deals with this subject.

 

[mmm4444bot]

so, whenever there is a fraction in the numerator do you invert it and put it in the denominator?

Not necessarily. It depends on what you're doing with the expression, afterwards. (It can also depend on the fraction itself; 1/8 ends up in the denominator as 8, but 3/8 in the numerator would not end up as 8/3 in the denominator -- unless you specifically desire that.)

In exercises like this one, which are designed to practice properties of exponents, you're not doing anything with the expressions afterwards. But they likely don't want to see negative exponents in the answers; they want you to demonstrate your understanding of the rule for changing a negative exponent to positive. (You could report 1/8 in the numerator as shown; there are no negative exponents. Your teacher might say, "Don't do that." If so, don't do it in that class. Technically, however, the 1/8 is not "wrong".)

Outside of such exercises for properties of exponents, we simplify expressions however we please (being mindful of any specific instructions), and this usually means simplifying steps or forms that fit nicely with what comes next. Maybe you switch things around more in some situations than others.

Here's a contrived example, where I would leave the 1/8 on top: If I recognized that I'm next multiplying by another ratio having a polynomial in the denominator whose coefficients were all multiples of 1/8, then I'd factor out 1/8 from the polynomial and cancel it with the 2^(-3) on top. :cool:

 

[sinx]

so, whenever there is a fraction in the numerator do you invert it and put it in the denominator?

you multiply the top and bottom by the same number, always (you are multiplying the fraction by 1)
^3/8/₄, you multiply ⁸/₈; =3/32