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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.
5a^6'b^-9
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c^-4'd^5
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. .
Found 2 solutions by Alan3354, Theo:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! 5a^6'b^-9
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c^-4'd^5
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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Use * (Shift 8) for multiply
5a^6*b^-9/(c^-4*d^5)
Don't remember rules if it's not necessary. Exponents don't move.
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5a^6*b^-9/(c^-4*d^5)
Multiply NUM and DEN by b^9
5a^6*b^0/(b^9*c^-4*d^5)
b^0 = 1
5a^6/(b^9*c^-4*d^5)
Multiply NUM and DEN by c^4
5a^6c^4/(b^9*c^0*d^5)
5a^6c^4/(b^9d^5)
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! the trick to dealing with negative exponents is very simple once you know it.
if the variable with the negative exponent is in the numerator, then place it in the denominator and make the exponent positive.
if the variable with the negative exponent is in the denominator, then place it in the numerator and make the exponent positive.
your expression is this:
5a^6*b^-9
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c^-4*d^5
i think you meant multiply so i used the * because that indicates multiplication.
b^-9 in the numerator goes to the denominator and becomes b^9.
c^-4 in the denominator goes to the numerator and becomes c^4
your expression becomes:
5a^6 * c^4
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b^9 * d^5
none of the variables in the numerator are the same as the variables in the denominator so no combining can be done.
the rules state that b^-x = 1/b^x.
this means that if it was in the numerator, it goes to the denominator and the exponent becomes positive.
the rules also state that 1/b^-x = b^x.
this means that if it was in the denominator, it goes to the numerator and the exponent becomes positive.
here's a link that discusses exponents that might help.
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut2_exp.htm
it all stems from the definition of negative exponents.
x^-a = 1/x^a
let's take a look at x^-a divided by y^-b
it looks like this:
x^-a
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y^-b
by the definition, x^&-a = 1/x^a and y^-b = 1/y^b
the expression becomes:
(1/x^a)
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(1/y^b)
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)
so you get (1/x^a) * (y^b)/1) which is equal to (y^b) / x^a)
you started with (x^-a) / (y^-b) and you ended up with (y^b) / x^a).
x^-a went to the denominator and became x^a.
y^-b went to the numerator and became y^b.
that's the rule that simplifies processing of these types of equations.
something else to consider.
suppose you have 5x^-4.
the x^-4 goes to the denominator.
the 5 stays because it doesn't have a negative exponent associated with it.
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.
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