hi this is my question:

(5^3)^2 X 25^2

i have no idea how to do this

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- Dec 8th 2009, 06:35 AMandyboy179indices (i think)
hi this is my question:

(5^3)^2 X 25^2

i have no idea how to do this - Dec 8th 2009, 06:44 AMe^(i*pi)
- Dec 8th 2009, 06:46 AMcraig
Do you want to solve $\displaystyle (5^3)^2 \times 25^2$ and get a numeric solution?

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

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

Following through with what I've done above, can you get $\displaystyle (5^3)^2 \times 25^2$ in terms of $\displaystyle 5^n$, where $\displaystyle n$ is just an integer (whole number) ? - Dec 8th 2009, 06:51 AMandyboy179
thank you very much for answering!!!

how would i work out 25^3 divided by 5^-4? - Dec 8th 2009, 06:54 AMcraig
Use what I said above to get $\displaystyle 25^3$ in terms of $\displaystyle 5^n$.

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

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

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. - Dec 8th 2009, 07:02 AMandyboy179
- Dec 8th 2009, 07:03 AMe^(i*pi)
- Dec 8th 2009, 07:03 AMcraig
- Dec 8th 2009, 07:07 AMandyboy179
5^-4= -25^-2 ???

- Dec 8th 2009, 07:08 AMe^(i*pi)
- Dec 8th 2009, 07:11 AMandyboy179
- Dec 8th 2009, 07:21 AMcraig
$\displaystyle \frac{25^3}{5^{-4}}$

$\displaystyle \frac{5^6}{5^{-4}} = 5^{6-(-4)} = $?? - Dec 8th 2009, 07:25 AMandyboy179
- Dec 8th 2009, 07:41 AMcraig
- Dec 8th 2009, 07:46 AMandyboy179
thankyou so much for all of the help!!!(Rock)