Question 77895
{{{(63^(-2))*(28^5)*7}}}
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The negative exponent can be changed to a positive exponent if the 63 is moved to a denominator.
This would make the problem:
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{{{(1/63^2)*(28^5)*7}}}
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Now factor the 63 into {{{(9*7)}}} and the 28 into {{{4*7}}}.  Substitute these into the
expression to get:
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{{{(1/(9*7)^2)*((4*7)^5)*7}}}
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by the power rule of exponents, the terms inside the parentheses each get raised to the
power associated with the brackets.  This converts the problem to:
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{{{1/(9^2*7^2)*(4^5)*(7^5)*7}}}
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in the numerator multiply the terms containing 7 by adding exponents (remember that 7
is the same as {{{(7^1)}}}) to get:
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{{{1/(9^2*7^2)* (4^5)*(7^6)}}}
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divide the {{{7^2}}} in the denominator into the {{{7^6}}} in the numerator by subtracting
the exponents to get {{{7^(6-2) = 7^4}}}.  This reduces the problem to:
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{{{1/(9^2)* (4^5)*(7^4) = ((4^5)*(7^4))/9^2}}}
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there are no more factors in the numerator that are common with factors in the denominator.
You could, if you wish, factor the 4 of the first term into the numerator into 2*2 and
the 9 in the denominator into 3*3 to get:
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{{{(((2*2)^5)*7^4)/(3*3)^2}}} which would expand to:
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{{{((2^5)*(2^5)*7^4)/(3^2*3^2) }}}
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but that doesn't accomplish much.  Or you could multiply everything out to get:
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{{{2458624/81}}}
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Hope this helps you to see how to work this problem.
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