Negative bases with exponents

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  • #1
DS2C
Going through a problem and and I keep getting it wrong and I'm not sure why.
In a part of the problem, the expression ##\left(-3\right)\left(-r^4\right)\left(-s^5\right)## comes up and the solution that it's giving me is ##-3r^4s^5##
Wouldn't the last factor be ##-s^5## since the power of a base with an odd exponent should be negative? Not sure if I'm tripped up somewhere but the book specifically states this, and then gives this solution.
 

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  • #2
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Going through a problem and and I keep getting it wrong and I'm not sure why.
In a part of the problem, the expression ##\left(-3\right)\left(-r^4\right)\left(-s^5\right)## comes up and the solution that it's giving me is ##-3r^4s^5##
Wouldn't the last factor be ##-s^5## since the power of a base with an odd exponent should be negative? Not sure if I'm tripped up somewhere but the book specifically states this, and then gives this solution.
There's a difference between ##-r^4## and ##(-r)^4## that you seem to be overlooking. The latter equals ##r^4##, which is the opposite sign of ##-r^4##. In the second and third factors, the bases are, respectively, r and s, not (-r) and (-s).
##(-3)(-r^4)(-s^5) = (-1)^3 \cdot 3r^4s^5 = -3r^4s^5##.
 
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  • #3
DS2C
To my understanding, ##-r^4## simply means "the negative of ##r^4##.
So ##-r^4=-\left(r\right)\left(r\right)\left(r\right)\left(r\right)##, and it will always be a negative no matter how many r factors there are and no matter if there is an even or odd amount.
Similarly, ##\left(-r\right)^4=\left(-r\right)\left(-r\right)\left(-r\right)\left(-r\right)=r^4##
However, if it were ##\left(-r\right)^5##, would this not be a negative since there is an odd exponent? Or am I just getting hung up on the even/odd exponent ordeal and way overthinking it? Will ##\left(-r\right)^4## and ##\left(-r\right)^5## BOTH have positive results?
For clarification, I plugged ##\left(-2\right)^5## into my calculator and it gave me -32. I then plugged in ##\left(-2\right)^6## and it gave me +64. So according to that train of thought, if the exponent is odd then the result will be negative and if the exponent is even then the result will be positive.
Kind of a dumb question, thank you for taking the time.
 
  • #4
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There is no (-r)5 involved.

Your expression is ##\left(-3\right)\left(-r^4\right)\left(-s^5\right) = (-3)(-(r^4))(-(s^5)) = (-3)(-1)(r^4)(-1)(s^5)## with additional brackets added to make the association clearer. If you simplify the last expression, you'll see that the exponents of r and s are irrelevant here.
 
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  • #5
DS2C
Ok I think I understand. I just used the ##\left(-r\right)^5## as an example not pulled from the actual expression.
I understand what youre both saying. But why does the calculator come up with something different, as in my last post?
 
  • #6
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For clarification, I plugged ##\left(-2\right)^5## into my calculator and it gave me -32. I then plugged in ##\left(-2\right)^6## and it gave me +64. So according to that train of thought, if the exponent is odd then the result will be negative and if the exponent is even then the result will be positive.
##(-2)^5## is the same as ##(-1)^5(2)^5##, which is -32. For the other expression, you have ##(-1)^6## times ##2^6##, or 64.
 
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  • #7
DS2C
Ok thanks for the help everyone.
 
  • #8
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The calculator probably operates step by step. You type in -2, then ^5, and it calculates (-2)^5.
 

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