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exponents - strange equation

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by 4meonly » Fri Nov 14, 2008 6:08 am
If 20/2^5 = 1/2^x + 1/2^y, where x and y are positive integers, what is the value of xy?
3
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Source: — Problem Solving |
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by cramya » Fri Nov 14, 2008 6:11 am
20/2^5 = 1/2 ^ X + 1/2 ^ Y

= 20/2^5 = 20/32 = 5/8

5/8 CAN BE EXPRESSED AS 1/8+4/8
= 1/8+ 1/2
= 1/2 ^ 3 + 1/2 ^ 1

x=3 y=1 or x=1 y=3 the product will be the same 3
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by jnellaz » Fri Nov 14, 2008 7:27 am
Cramya - that was very clever. I believe I know the basic rules very well, but I keep seeing problems that (at least to me) seem as if you need an imaginative sense to solve and a keen eye to avoid going down the wrong path.

Can you suggest any math resources or books that can assist in learning these harder types of fraction/exponent problems that require a bit more ingenuity? Thanks again!
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by cramya » Fri Nov 14, 2008 7:43 am
Jnellaz,
My humble opinion would be for u to use the search option on this forum and search for exponents. You will see several problem solved and that may provide you more insights in trying various things out.

To tell u the truth I learnt more about manipulating exponents from this forum that any other text/resource.

A general rule is to get the bases on lhs the same as rhs which is what I did. breaking up of 5/8 = 1/8 + 4/8 is to get the bases 1/2 on both sides.

Hope this helps! Let me know if u have further questions.

Regards,
Cramya
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