1. ## evaluating numerical expression

Hello forum, Vaironxxrd here.

I have a question about exponents. If I have,

$(-9)(-9)^3$ Would that be $(-9)^4$ or $(-9^4)$.

If possible can you guys provide other examples?

2. ## Re: evaluating numerical expression

It's indeed $(-9)^4$. For example, consider $(-2)^4\cdot (-2)^2$ which is offcourse $(-2)^6=64$ and $-2^6=-(2^6)=-64$ therefore there're not equal.
So it's important to use brackets!

3. ## Re: evaluating numerical expression

Originally Posted by Siron
It's indeed $(-9)^4$. For example, consider $(-2)^4\cdot (-2)^2$ which is offcourse $(-2)^6=64$ and $-2^6=-(2^6)=-64$ therefore there're not equal.
So it's important to use brackets!
Finally, I now understand it (-9)^3 = -729 is going to be something negative since the exponent is odd and I just do the normal exponent.
9^2 is positive = 81.

Now what is the difference between (-9^2) and (-9)^2

4. ## Re: evaluating numerical expression

Originally Posted by vaironxxrd
Finally, I now understand it (-9)^3 = -729 is going to be something negative since the exponent is odd and I just do the normal exponent.
9^2 is positive = 81.

Now what is the difference between (-9^2) and (-9)^2
Exponents come before multiplication in the order of operations and a minus sign in front is multiplying by -1.

However, brackets come before exponents so if the multiplication is done inside the bracket it too is affected by the exponents.

$(-9)^2 = -9 \times -9 = 81 \text{ OR } (-9)^2 = (-1 \times 9)^2 = (-1)^2 \times 9^2 = 81$

$-9^2 = -1 \times 9^2 = -81$ - the brackets here are unnecessary and only serve to complicate matters IMO.

5. ## Re: evaluating numerical expression

Originally Posted by e^(i*pi)
Exponents come before multiplication in the order of operations and a minus sign in front is multiplying by -1.

However, brackets come before exponents so if the multiplication is done inside the bracket it too is affected by the exponents.

$(-9)^2 = -9 \times -9 = 81 \text{ OR } (-9)^2 = (-1 \times 9)^2 = (-1)^2 \times 9^2 = 81$

$-9^2 = -1 \times 9^2 = -81$ - the brackets here are unnecessary and only serve to complicate matters IMO.
Thanks allot Pi!

6. ## Re: evaluating numerical expression

Originally Posted by vaironxxrd
Now what is the difference between (-9^2) and (-9)^2
Remember BIDMAS (Brackets, Indices, Division, Multiplication, Addition, Subtraction).
The order of operands is very important, and this is where you're having trouble.

-9 is essentially like saying 0 - 9, you just omit the 0.
Now, from BIDMAS, we know that everything in brackets is evaluated first, and that indices/multiplication is evaluated before subtraction.

So (-9^2) is like saying (0 - 9^2). You evaluate the index first, then you subtract the result from 0.
0 - (9)(9) = 0 - 81 = -81

Whereas, in (-9)^2 you evaluate everything in brackets first, so (0 - 9)^2 = (-9)(-9) = 81, because multiplying an even number of negatives together results in a positive.

I hope that was clear enough.

7. ## Re: evaluating numerical expression

Originally Posted by FrameOfMind
Remember BIDMAS (Brackets, Indices, Division, Multiplication, Addition, Subtraction).
The order of operands is very important, and this is where you're having trouble.

-9 is essentially like saying 0 - 9, you just omit the 0.
Now, from BIDMAS, we know that everything in brackets is evaluated first, and that indices/multiplication is evaluated before subtraction.

So (-9^2) is like saying (0 - 9^2). You evaluate the index first, then you subtract the result from 0.
0 - (9)(9) = 0 - 81 = -81

Whereas, in (-9)^2 you evaluate everything in brackets first, so (0 - 9)^2 = (-9)(-9) = 81, because multiplying an even number of negatives together results in a positive.

I hope that was clear enough.
To not start a new thread can you tell me if I correctly solved this problem?

$\frac{5x^3y^9}{20x^2y^-^2}$ Solution: $\frac{x&y^1^1}{4}$

Is this right?

8. ## Re: evaluating numerical expression

Originally Posted by vaironxxrd
To not start a new thread can you tell me if I correctly solved this problem?

$\frac{5x^3y^9}{20x^2y^-2}$ Solution: $\frac{x&y^1^1}{4}$

Is this right?
if that is $y^{-2}$ in the denominator, then yes, it's correct.

In future, go ahead and start a new problem with a new thread.

9. ## Re: evaluating numerical expression

Yeah sorry for that. Does it matter how many threads I start? I feel like it might be annoying.

10. ## Re: evaluating numerical expression

Originally Posted by vaironxxrd
Yeah sorry for that. Does it matter how many threads I start? I feel like it might be annoying.
a single thread with multiple problems all strung out is more annoying.