SOLUTION: (2 to the power of 6 and 4 over 2 to the power of 2 a.2 to the power 22 b.2 to the power 12 c.2 to the power of 8 d.2 to the power 5

Algebra ->  Exponents -> SOLUTION: (2 to the power of 6 and 4 over 2 to the power of 2 a.2 to the power 22 b.2 to the power 12 c.2 to the power of 8 d.2 to the power 5      Log On


   

Question 596259: (2 to the power of 6 and 4 over 2 to the power of 2
a.2 to the power 22
b.2 to the power 12
c.2 to the power of 8
d.2 to the power 5
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
I think you are saying that the problem is to simplify:
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%28%282%5E6%29%5E4%29%2F2%5E2
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If I have interpreted your problem correctly, here's how you would do it.
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First apply the power rule of exponents to the numerator. This rule says that when you have an exponent term raised to a power, you multiply the two exponents. (What you really have in the numerator is 2^6 times 2^6 times 2^6 times 2^6 and you can simplify this by adding exponents to get 2^24. This is done easier by just multiplying the exponent 6 times the exponent 4 to get the same answer of 2^24.) So the problem becomes:
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2%5E24%2F2%5E2
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To divide terms with the same base (in this problem the base is 2) you just subtract the exponent in the denominator from the exponent in the numerator to get:
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2%5E%2824-2%29
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And after subtracting the exponents you get the answer of:
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2%5E22
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This would be answer a in the list of choices you were given.
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Again, this answer assumes that I interpreted your problem correctly. If I had the problem correct at the start, then the answer I got is the correct one. If I did not start with the correct problem, then you will need to clarify the problem statement and post it again.
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Hope this helps you understand the problem a little more, and how you can work it to a solution.
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