document.write( "Question 965052: I'm doing negative exponents and they're killing me. I'm trying to understand the concept,especially this certain question. I hope you can help me.\r
\n" ); document.write( "\n" ); document.write( "5a^6'b^-9
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\n" ); document.write( "\n" ); document.write( "The answer key has the c^-4 joining the 5 (5c^4) as well as a^-6, b^-9 moving to the denominator with d^5. I hope you can understand this email and explain to me why these exponents moved as they did. .
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Algebra.Com's Answer #589828 by Theo(13342)\"\" \"About 
You can put this solution on YOUR website!
the trick to dealing with negative exponents is very simple once you know it.\r
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\n" ); document.write( "\n" ); document.write( "if the variable with the negative exponent is in the numerator, then place it in the denominator and make the exponent positive.\r
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\n" ); document.write( "\n" ); document.write( "if the variable with the negative exponent is in the denominator, then place it in the numerator and make the exponent positive.\r
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\n" ); document.write( "\n" ); document.write( "your expression is this:\r
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\n" ); document.write( "\n" ); document.write( "5a^6*b^-9
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\n" ); document.write( "\n" ); document.write( "i think you meant multiply so i used the * because that indicates multiplication.\r
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\n" ); document.write( "\n" ); document.write( "b^-9 in the numerator goes to the denominator and becomes b^9.\r
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\n" ); document.write( "\n" ); document.write( "c^-4 in the denominator goes to the numerator and becomes c^4\r
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\n" ); document.write( "\n" ); document.write( "your expression becomes:\r
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\n" ); document.write( "\n" ); document.write( "5a^6 * c^4
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\n" ); document.write( "\n" ); document.write( "none of the variables in the numerator are the same as the variables in the denominator so no combining can be done.\r
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\n" ); document.write( "\n" ); document.write( "the rules state that b^-x = 1/b^x.\r
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\n" ); document.write( "\n" ); document.write( "this means that if it was in the numerator, it goes to the denominator and the exponent becomes positive.\r
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\n" ); document.write( "\n" ); document.write( "the rules also state that 1/b^-x = b^x.\r
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\n" ); document.write( "\n" ); document.write( "this means that if it was in the denominator, it goes to the numerator and the exponent becomes positive.\r
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\n" ); document.write( "\n" ); document.write( "here's a link that discusses exponents that might help.\r
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\n" ); document.write( "\n" ); document.write( "http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut2_exp.htm\r
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\n" ); document.write( "\n" ); document.write( "it all stems from the definition of negative exponents.\r
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\n" ); document.write( "\n" ); document.write( "x^-a = 1/x^a\r
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\n" ); document.write( "\n" ); document.write( "let's take a look at x^-a divided by y^-b\r
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\n" ); document.write( "\n" ); document.write( "it looks like this:\r
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\n" ); document.write( "\n" ); document.write( "x^-a
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\n" ); document.write( "\n" ); document.write( "by the definition, x^&-a = 1/x^a and y^-b = 1/y^b\r
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\n" ); document.write( "\n" ); document.write( "the expression becomes:\r
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\n" ); document.write( "\n" ); document.write( "(1/x^a)
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\n" ); document.write( "\n" ); document.write( "this is equivalent to (1/x^a) * (y^b/1) by the law of multiplication that states that (a/b) / (c/d) = (a/b) * (d/c)\r
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\n" ); document.write( "\n" ); document.write( "so you get (1/x^a) * (y^b)/1) which is equal to (y^b) / x^a)\r
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\n" ); document.write( "\n" ); document.write( "you started with (x^-a) / (y^-b) and you ended up with (y^b) / x^a).\r
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\n" ); document.write( "\n" ); document.write( "x^-a went to the denominator and became x^a.
\n" ); document.write( "y^-b went to the numerator and became y^b.\r
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\n" ); document.write( "\n" ); document.write( "that's the rule that simplifies processing of these types of equations.\r
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\n" ); document.write( "\n" ); document.write( "something else to consider.\r
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\n" ); document.write( "\n" ); document.write( "suppose you have 5x^-4.\r
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\n" ); document.write( "\n" ); document.write( "the x^-4 goes to the denominator.
\n" ); document.write( "the 5 stays because it doesn't have a negative exponent associated with it.\r
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\n" ); document.write( "\n" ); document.write( "however, (5x)^-4 becomes 1/(5x)^-4 because now the 5 and the x have the negative exponent associated with them because the 5x is surrounded by the parentheses.\r
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