Question 75556
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{{{(b^2)^(-6)/b^(-4)}}}
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First, we'll convert this to have only positive exponents, and then we'll simplify.
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Let's get the numerator to have a positive exponent.  Begin by using the power rule for
exponents.  This rule says that if you have a term with an exponent that is then raised
to another exponent, you multiply the two exponents.  So:
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{{{(b^2)^(-6)}}} becomes {{{b^(2*(-6)) = b^(-12)}}}.  But a base raised to a negative 
number is equal to that same base to a positive number and then divided into 1. So the 
numerator {{{b^(-12)}}} becomes {{{1}}} divided by {{{b^(12)}}} which is written as:
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{{{1/(b^12)}}}
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Now work on the denominator. {{{(b)^(-4)}}} just falls under the rule to divide 1 by the
same base with a positive exponent.  So the denominator becomes:  
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{{{1/(b^(4))}}}
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Now putting the numerator and denominator together you get:
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{{{(1/(b^12))/(1/(b^(4)))}}}
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Now recall that a long time ago you learned that when you divide by a fraction, you invert
the divisor and multiply the dividend by it.  When you set that up, you get:
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{{{(1/(b^12))*((b^4)/1)}}}
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When you multiply the numerators and denominators of both these terms you get:
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{{{(1*b^4)/((b^12)*1)}}}
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which simplifies to:
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{{{b^4/b^12}}}
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and you perform this division by subtracting the exponent in the denominator from the
exponent in the numerator to get: {{{b^(4-12) = b^-8}}}.  But there is a negative exponent
and the problem calls for all positive exponents.  So just divide the base with a 
positive exponent into 1 to get: {{{1/b^8}}} and that is your answer.
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Hope this helps you to understand the workings of negative exponents.