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Alex787 [66]
3 years ago
12

Please help me with this , I will give 30 POINTS !.

Mathematics
1 answer:
Katen [24]3 years ago
8 0
<h3>Answer:  B) 16</h3>

======================================================

Explanation:

Through trial and error, we can set up these equations

  • 4^5 = 1024
  • 2^10 = 1024
  • 32^2 = 1024

This shows that choices A, C and D are eliminated. This is because 5, 10 and 32 all can work as either x or y in x^y = 1024, and that the other value is some other integer.

On the other hand, x = 16 doesn't work and neither does y = 16 if we wanted the other value to be an integer.

Solving x^16 = 1024 leads to approximately x = 1.5422 when we focus on the positive real number solution. This isn't an integer, so x^16 = 1024 doens't have any integer solutions.

Solving 16^y = 1024 leads to a similar situation as described in the previous paragraph. It doesn't have any integer solutions either (instead the value is y = 2.5)

So we have a few ways of arriving at why choice B is the final answer.

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