# Indices - Numbers as a product of prime numbers

I've checked the internet which only provides basic $x^2 \times x^3 = x^5$ information and have concluded that I need resort to a Q & A website. The basics of indices are fine for me, but it's when it gets to fractions in the powers and being told to do write them as "as a product of powers of prime numbers" is when I fall short.

For example: $25^2 \times 10^{1/2}$, I think I could square $25$ and multiply that by the square root of $10$ but I think that is a too literal approach. I'm fine with stuff like: $15^3 = 5^3 \times 3^3$, just those darn fractions.

Any help would be appreciated, this must seem like a rather primitive question for such a high level community, but this has been grinding on me for a couple of days now and I really need help.

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Just factorise all the bases of powers in your expression as a product of primes and then collect the powers of the same prime together. So for your example:

$25^2 * 10^{1/2} = (5^2)^2 * (2*5)^{1/2} = 5^4*2^{1/2}*5^{1/2} = 5^{9/2}*2^{1/2}$.

(I've used $*$ for multiplication.)

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This answer was a bit more comprehensible at my level of Mathematics, many thanks! Will accept when I can. – PwnageAtPwn Feb 12 '13 at 19:56

We write the expression as the product of powers of primes:

\begin{align} 25^2 \times 10^{1/2} &= \left((5)^2\right)^2 \times (5\times 2)^{1/2} \\ \\ &= 5^4\times 5^{1/2} \times 2^{1/2} \\ \\ &= 2^{1/2}\times 5^{4 + \large\frac12} \\ \\ &= 2^{1/2}\times 5^{9/2} \end{align}

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Haha, snap! Except you were first. – Tara B Feb 12 '13 at 19:53
+1 neat ${}{}{}{}$ – Babak S. Feb 24 '13 at 12:46
@Babak Thanks for that! :-) – amWhy Feb 24 '13 at 13:00