anonymous
  • anonymous
@Osako
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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phi
  • phi
if you write 1 as x^0 and subtract exponents (top exponent minus bottom exponent) what do you get ?
phi
  • phi
in other words simplfy \[ \frac{ x^0}{x^{-1} }\] using the rule "subtract exponents" like you did on the earlier problem

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anonymous
  • anonymous
I'm confused
anonymous
  • anonymous
0-(-1)?
phi
  • phi
yes
anonymous
  • anonymous
so -1 is the answer? why can you just get rid of the 1 for a 0
phi
  • phi
? are you saying 0 - (-1) is -1 ?
anonymous
  • anonymous
1*
anonymous
  • anonymous
sorry, lol
phi
  • phi
so 1 is the exponent. the base is x. the answer is x^1 or just x so notice \[ \frac{1}{x^{-1}} = x^1 \]
phi
  • phi
notice this: you know x divided by itself is 1 \[ \frac{x}{x} = 1\] If we use the "subtract exponents" rule on \[ \frac{x^1}{x^1} = \] you get ?
phi
  • phi
you get an exponent of 1-1 = 0 and a base of x you get \[ \frac{x}{x}= x^0 \] but x/x is 1, so we find \[ 1 = x^0\]
anonymous
  • anonymous
0?
anonymous
  • anonymous
thank you
phi
  • phi
we can use the rule to show \[ \frac{1}{x} = \frac{x^0}{x^1} \] now use the subtract rule to get \[ \frac{x^0}{x^1} = x^{0-1} = x^{-1} \] or \[ \frac{1}{x} = x^{-1} \]
anonymous
  • anonymous
Could it equal 1?
anonymous
  • anonymous
It should based on the question I have
phi
  • phi
Could it equal 1 ? what is "it" ?
anonymous
  • anonymous
The answer to the problem
phi
  • phi
\[ \frac{1}{x^{-1}} = x \] that is true for any x (except x being 0) for example \[ \frac{1}{2^{-1}} = 2 \] a negative exponent means you can "flip" the number, and make the exponent positive.

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