evaluating numerical expression

• Oct 8th 2011, 07:47 AM
vaironxxrd
evaluating numerical expression
Hello forum, Vaironxxrd here.

I have a question about exponents. If I have,

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

If possible can you guys provide other examples?
• Oct 8th 2011, 07:57 AM
Siron
Re: evaluating numerical expression
It's indeed $\displaystyle (-9)^4$. For example, consider $\displaystyle (-2)^4\cdot (-2)^2$ which is offcourse $\displaystyle (-2)^6=64$ and $\displaystyle -2^6=-(2^6)=-64$ therefore there're not equal.
So it's important to use brackets!
• Oct 8th 2011, 09:54 AM
vaironxxrd
Re: evaluating numerical expression
Quote:

Originally Posted by Siron
It's indeed $\displaystyle (-9)^4$. For example, consider $\displaystyle (-2)^4\cdot (-2)^2$ which is offcourse $\displaystyle (-2)^6=64$ and $\displaystyle -2^6=-(2^6)=-64$ 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
• Oct 8th 2011, 09:58 AM
e^(i*pi)
Re: evaluating numerical expression
Quote:

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.

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

$\displaystyle -9^2 = -1 \times 9^2 = -81$ - the brackets here are unnecessary and only serve to complicate matters IMO.
• Oct 8th 2011, 10:02 AM
vaironxxrd
Re: evaluating numerical expression
Quote:

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.

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

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

Thanks allot Pi!
• Oct 8th 2011, 10:06 AM
FrameOfMind
Re: evaluating numerical expression
Quote:

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.
• Oct 9th 2011, 07:34 AM
vaironxxrd
Re: evaluating numerical expression
Quote:

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?

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

Is this right?
• Oct 9th 2011, 07:41 AM
skeeter
Re: evaluating numerical expression
Quote:

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

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

Is this right?

if that is $\displaystyle y^{-2}$ in the denominator, then yes, it's correct.

In future, go ahead and start a new problem with a new thread.
• Oct 9th 2011, 07:43 AM
vaironxxrd
Re: evaluating numerical expression
Yeah sorry for that. Does it matter how many threads I start? I feel like it might be annoying.
• Oct 9th 2011, 07:55 AM
skeeter
Re: evaluating numerical expression
Quote:

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.