SOLUTION: What expression raised to the fourth power is 81x12y8z16? will it be polynomials with more than one variable

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Question 74023: What expression raised to the fourth power is 81x12y8z16?
will it be polynomials with more than one variable

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given: %2881%29%2A%28x%5E12%29%2A%28y%5E8%29%2A%28z%5E16%29
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To find the term that, when raised to the fourth power, results in the above expression,
you can just take the fourth root of the above expression. One way of writing the fourth root
is to just raise the above expression to the 1%2F4 power. Then you can use the rules
of exponents. When you raise the above to the 1%2F4 power, the expression becomes:
.
%28%2881%29%2A%28x%5E12%29%2A%28y%5E8%29%2A%28z%5E16%29%29%5E%281%2F4%29
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next you apply the 1%2F4 power to each of the factors in the given expression to get:
.

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The first factor of 81 has a fourth root of 3 because 3*3*3*3 = 81. So let's replace 81%5E%281%2F4%29
by 3. When you do the expression is simplified to:
.
3%2A%28x%5E12%29%5E%281%2F4%29%2A%28y%5E8%29%5E%281%2F4%29%2A%28z%5E16%29%5E%281%2F4%29
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Then each of the other terms is raised to the 1%2F4 power by multiplying its exponent by 1%2F4.
So:
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%28x%5E12%29%5E%281%2F4%29+=+x%5E%2812%2F4%29+=+x%5E3 and
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%28y%5E8%29%5E%281%2F4%29+=+y%5E%288%2F4%29+=+y%5E2 and
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%28z%5E16%29%5E%281%2F4%29+=+z%5E%2816%2F4%29+=+z%5E4
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Multiplying all these answers together results in:
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3%2Ax%5E3%2Ay%5E2%2Az%5E4
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And that's the answer. If you reverse the process and raise the answer to the fourth
power, you get back to the original expression.
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Hope this helps you to see how the rules of exponents can be applied to a problem such
as this. You should be particularly aware that an exponent of 1%2F2 is equivalent
to taking the square root, and exponent of 1%2F3 is equivalent to taking the cube root,
and, as in this case, an exponent of 1%2F4 is equivalent to finding the fourth root ...
and so on.