# Thread: Exponent Multiplication and Division (again)

1. ## Exponent Multiplication and Division (again)

For the following expression I am having difficulty understanding why one cannot proceed with the rule $a^na^m=a^{n+m}$ with a -9 from $(3xy^4)^{-2}$, instead of first converting it to a fraction first e.g $(\frac{1}{9}x^{-2}y^{-8})(144x^4y^2)$

Expression

$(3xy^4)^{-2}(12x^2y)^2$

= $(\frac{1}{9}x^{-2}y^{-8})(144x^4y^2)= 16x^2y^{-6}$
Thats as far as I will simplify.

Why cannot this read $(-9x^{-2}y^{-8})(144x^4y^2) ?$

2. I think the technique that you are trying to apply is moving a negative power from the denominator, and while you are on the right track, what you have done is incorrect.

$
\frac{1}{9}=9^{-1}
$

$
\frac{1}{9}\ne-9
$

It is easy to get scooped into this as it is custom to change, for example, $2y^{-2}$ to $\frac{2}{y^2}$. But it is extremely important to remember that if you are going to do this with a constant with no visible exponent, it really looks like this.

$
\frac{1}{9}=\frac{1}{9^1}=9^{-1}
$

So that if you bring that 9 up, it's exponent value of 1 becomes negative, not the 9 itself.

3. Also one other thing to keep in mind if you are unsure of a method of simplification, or want to double check your answer. Remember that simplifications are aesthetic and never change the actual value of the function. If you simplify A to B to C. The following must ALWAYS be true. $A=B=C$.

With that in mind, you can check the value of any given equation relative to others by picking arbitrary values and plugging them in to get a value. This value should be equal to any equations preceeding it in the simplification process and any coming after it.

For the sake of example, I am choosing $x=5$ and $y=10$.

First equation - The $\frac{1}{9}$ equation.
$(\frac{1}{9}x^{-2}y^{-8})(144x^4y^2)$
= $(\frac{1}{9}5^{-2}10^{-8})(144(5)^4(10)^2)$
= $0.0004$

Second Equation - The $9^{-1}$ equation.
$(9^{-1}x^{-2}y^{-8})(144x^{4}y^{2})$
= $(9^{-1}(5)^{-2}(10)^{-8})(144(5)^{4}(10)^{2})$
= $0.0004$

Third equation - Comparing the above to the final simplified $16x^2y^{-6}$
$16x^2y^{-6}$
= $16(5)^2(10)^{-6}$
= $0.0004$

Finally, we will plug these values into the $-9$ equation and see whats comes out.
$(-9x^{-2}y^{-8})(144x^{4}y^2)$
= $(-9(5)^{-2}(10)^{-8})(144(5)^{4}(10)^2)$
= $-0.0324$

As you can see the method of simplification you used is not valid as it causes the value of the equation with the same x and y values to change. Does this help?

4. Unfortunately Kasper 1/9 is correct according to the book in this case.

They solve it as follows
$(3xy^4)^{-2}(12x^2y)^2$
$(\frac{1}{9}x^{-2}y^{-8})(144x^4y^2)= 16x^2y^{-6}=\frac{16x^2}{y^6}$

Again im not sure why the 1/9 does not remain as -9.

5. Originally Posted by allyourbass2212
Unfortunately Kasper 1/9 is correct according to the book in this case.

They solve it as follows
$(3xy^4)^{-2}(12x^2y)^2$
$(\frac{1}{9}x^{-2}y^{-8})(144x^4y^2)= 16x^2y^{-6}=\frac{16x^2}{y^6}$

Again im not sure why the 1/9 does not remain as -9.
No, you read my post wrong, I am saying that $\frac{1}{9}$ is correct and that $-9$ is not correct. My first post correlates to the solution you provided by the following statement:

$3^{-2}=9^{-1}=\frac{1}{9}\ne-9$

The $\frac{1}{9}$ never was a $-9$, it could not have been because the two are not equivalent.
Read over my first post about how negative exponents work, let me know if you still don't understand.

6. Ive read over your post but I am still a little confused.

For instance

$3^{-2}=9^{-1}=\frac{1}{9}\ne-9$

So it appears any number to a negative exponent you can rewrite as $n^{-1}$ rather than $-n$.

For instance
$4^{-2}=16^{-1}$
$7^{-2}=49^{-1}$

Is this correct?

If so that would allow us to use $a^{-n} = \frac{1}{a^n}$, in which case the $16^{-1}$would equal $\frac{1}{16}$ and $49^{-1}$ would equal $\frac{1}{49}$.

Is my logic correct?

7. Originally Posted by allyourbass2212
Ive read over your post but I am still a little confused.

For instance

$3^{-2}=9^{-1}=\frac{1}{9}\ne-9$

So it appears any number to a negative exponent you can rewrite as $n^{-1}$ rather than $-n$.

For instance
$4^{-2}=16^{-1}$
$7^{-2}=49^{-1}$

Is this correct?

If so that would allow us to use $a^{-n} = \frac{1}{a^n}$, in which case the $16^{-1}$would equal $\frac{1}{16}$ and $49^{-1}$ would equal $\frac{1}{49}$.

Is my logic correct?

Very correct

8. Originally Posted by allyourbass2212
Ive read over your post but I am still a little confused.

For instance

$3^{-2}=9^{-1}=\frac{1}{9}\ne-9$

So it appears any number to a negative exponent you can rewrite as $n^{-1}$ rather than $-n$.

For instance
$4^{-2}=16^{-1}$
$7^{-2}=49^{-1}$

Is this correct?

If so that would allow us to use $a^{-n} = \frac{1}{a^n}$, in which case the $16^{-1}$would equal $\frac{1}{16}$ and $49^{-1}$ would equal $\frac{1}{49}$.

Is my logic correct?
Exactly, and another thing to try to keep these rules fresh in your mind. Just stick em in your calculator. Expressions with equivalent values are just that, equivalent, and therefore simplifications of each other.

For example, let's consider the statement:

$4^{-2}=16^{-1}=\frac{1}{16}\ne-16$

Just plug each one into your calculator to find what is equivalent and what isn't. Like a "One of these does not belong." game, .

i.e.
$4^{-2}=0.0625$

$16^{-1}=0.0625$

$\frac{1}{16}=0.0625$

$-16=-16$

Also, just to clarify one piece of your post.

So it appears any number to a negative exponent you can rewrite as $n^{-1}$ rather than $-n$.
This is correct, but I would like to reinforce that it is not a *can*. If you are simplifying it this way, that is exactly what needs to be done because it is never equal to -n. Any number to a negative exponent is never equal to just ignoring the negative, applying the power, and then bringing the negative down to make the entire expression negative.

In other words, to reapply the concepts above, $2^{-3}$ is never equal to $-8$.

9. Thanks again guys!