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Exponents and Roots

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corinna
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Exponents and Roots [#permalink] New post   28 Oct 2011, 03:35
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Simplify the following fraction:

2^21/4^4(4^5-4^3)



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corinna
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Re: Exponents and Roots [#permalink] New post   28 Oct 2011, 05:12
answer is supposed to be: 2^7/15?!
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corinna
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Re: Exponents and Roots [#permalink] New post   28 Oct 2011, 10:55
Can anybody explain the way of calculation? That would be really great!
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Re: Exponents and Roots [#permalink] New post   28 Oct 2011, 16:27
Hi Corinna,

Here is my working out.

First, let's simplify the bottom expression.

4^4 (4^5 -4^3)

Expand

4^9 - 4^7

Take out common factor

4^7 (4^2 - 1)

= 4^7 (15)

Turn the 4 into a base of 2, which is 2^2

So (2^2)^7 ( 15)

= 2^14 (15)

Now we can cancel out the 2^14 by the 2^21 on the top so the top becomes 2^7

So the answer is

2^7/15

I hope that makes sense.
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Re: Exponents and Roots [#permalink] New post   28 Oct 2011, 18:28
This question is a bit silly without answer choices as it's not clear when the simplification is finished, but anyways:

\(\frac{2^{21}}{4^4*(4^5-4^3)}\)

\(\frac{2^{21}}{2^8*(2^{10}-2^6)}\)

\(\frac{2^{21}}{2^{14}*(2^4-1)}\)

\(\frac{2^{7}}{2^4-1}\)

\(\frac{2^{7}}{15}\)

\(\frac{128}{15}\)



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Re: Exponents and Roots [#permalink]
28 Oct 2011, 18:28
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