indices (i think)

**andyboy179** · December 8th 2009, 06:35 AM

hi this is my question:

(5^3)^2 X 25^2

i have no idea how to do this

**e^(i*pi)** · December 8th 2009, 06:44 AM

Originally Posted by andyboy179

hi this is my question:

(5^3)^2 X 25^2

i have no idea how to do this

$5^6 \times 5^4 = 5^{10} = 9 765 625$

**craig** · December 8th 2009, 06:46 AM

Originally Posted by andyboy179

hi this is my question:

(5^3)^2 x 25^2

i have no idea how to do this

Do you want to solve $(5^3)^2 \times 25^2$ and get a numeric solution?

$(5^3)^2 = 5^{3\times 2} = 5^6$

Also note that $25 = 5^2$ , this means that $25^2 = (5^2)^2 = 5^4$ .

Following through with what I've done above, can you get $(5^3)^2 \times 25^2$ in terms of $5^n$ , where $n$ is just an integer (whole number) ?

**andyboy179** · December 8th 2009, 06:51 AM

thank you very much for answering!!!

how would i work out 25^3 divided by 5^-4?

**craig** · December 8th 2009, 06:54 AM

Originally Posted by andyboy179

thank you very much for answering!!!

how would i work out 25^3 divided by 5^-4?

Use what I said above to get $25^3$ in terms of $5^n$ .

Also notice that $5^n \times 5^m = 5^{n+m}$ .

This also applies for division, $\frac{5^n}{5^m} = 5^{n-m}$ .

Another thing, do not just post questions and hope for the answer, show us your working so far and where you have got stuck, then we can help you.

**andyboy179** · December 8th 2009, 07:02 AM

Originally Posted by craig

Use what I said above to get $25^3$ in terms of $5^n$ .

Also notice that $5^n \times 5^m = 5^{n+m}$ .

This also applies for division, $\frac{5^n}{5^m} = 5^{n-m}$ .

Another thing, do not just post questions and hope for the answer, show us your working so far and where you have got stuck, then we can help you.

would 25^3= 5^6?

and could you explain what to do on 5^-4???

**e^(i*pi)** · December 8th 2009, 07:03 AM

Originally Posted by andyboy179

would 25^3= 5^6?

Yes

and could you explain what to do on 5^-4???

Recall that for $a^{-b} = \frac{1}{a^b}$

**craig** · December 8th 2009, 07:03 AM

Originally Posted by andyboy179

would 25^3= 5^6?

Correct!

Originally Posted by andyboy179

and could you explain what to do on 5^-4???

Use the fact that $\frac{5^n}{5^m} = 5^{n-m}$ , just remember that it's a minus 4 you are using, not 4.

**andyboy179** · December 8th 2009, 07:07 AM

5^-4= -25^-2 ???

**e^(i*pi)** · December 8th 2009, 07:08 AM

Originally Posted by andyboy179

5^-4= -25^-2 ???

Not quite, the exponent is fine but consider the sign on $25^{-2}$

**andyboy179** · December 8th 2009, 07:11 AM

Originally Posted by e^(i*pi)

Not quite, the exponent is fine but consider the sign on $25^{-2}$

so would i just be -25^2

**craig** · December 8th 2009, 07:21 AM

$\frac{25^3}{5^{-4}}$

$\frac{5^6}{5^{-4}} = 5^{6-(-4)} =$ ??

**andyboy179** · December 8th 2009, 07:25 AM

Originally Posted by craig

$\frac{25^3}{5^{-4}}$

$\frac{5^6}{5^{-4}} = 5^{6-(-4)} =$ ??

what so i leave 56-4 the same and then do the divide part?

would it = 5^10

**craig** · December 8th 2009, 07:41 AM

Originally Posted by andyboy179

what so i leave 56-4 the same and then do the divide part?

would it = 5^10

Yes. When multiply or dividing just remember the two rules above. It doesn't matter what the base numbers are, (5 or 25), as long as they are the same.

Hope this has helped, be sure to post again if there's anything you stuck on

**andyboy179** · December 8th 2009, 07:46 AM

thankyou so much for all of the help!!!

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