Spring98
  • Spring98
Medals!! Simplify 12^3/12^7.
Mathematics
katieb
  • katieb
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javalos01
  • javalos01
ik
javalos01
  • javalos01
it
Spring98
  • Spring98
?

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rishavraj
  • rishavraj
\[\frac{ 1 }{ a^2 } = a^{-2}~~and~~ a^x \times a^y = a^{x+ y}\]
Spring98
  • Spring98
so how do i do it?
rishavraj
  • rishavraj
\[\frac{ 1 }{ 12^{7} } = 12^{-7}\]
CGGURUMANJUNATH
  • CGGURUMANJUNATH
a^x / a^y =a^(x-y)
Jhannybean
  • Jhannybean
\[\frac{12^3}{12^7} = 12^{3-7} = 12^?\]
CGGURUMANJUNATH
  • CGGURUMANJUNATH
a=12,x=3,y=7
Jhannybean
  • Jhannybean
\[\large 12^{-?} \iff \frac{1}{12^?}\]
Spring98
  • Spring98
4
rishavraj
  • rishavraj
so it is \[\frac{ 1 }{ 12^4 } \] now wht??
rishavraj
  • rishavraj
remember \[\frac{ 1 }{ a } = a^{-1}\]
Spring98
  • Spring98
oh so would it be - 1/12^4?
anonymous
  • anonymous
12^4
anonymous
  • anonymous
that would be you answer
rishavraj
  • rishavraj
rishavraj
  • rishavraj
@Spring98 u done???
anonymous
  • anonymous
yah why not?
anonymous
  • anonymous
there's no negative exponents .
rishavraj
  • rishavraj
@yinkim52001 remember \[\frac{ 1 }{ a } = a^{-1}\]
Jhannybean
  • Jhannybean
You never leave your answers with negative exponents unless it's called for. Therefore \[\frac{1}{12^4} = \frac{1}{12\cdot 12\cdot 12\cdot 12}\] Whatever that is is your answer.
Spring98
  • Spring98
12^4 is the answer right?
Jhannybean
  • Jhannybean
No.
anonymous
  • anonymous
no its 1/12^4
anonymous
  • anonymous
i see why its 1/12^4
rishavraj
  • rishavraj
@Jhannybean its asking to just simplfy it ..... so i think we may just write in exponential form
Jhannybean
  • Jhannybean
You can have various simplifications @rishavraj either in positive exponential form, fully expanded form, or even negative exponent form. They would all be correct.
Spring98
  • Spring98
what is the answer then ? 1/12^4?
anonymous
  • anonymous
\[\frac{ 1 }{ 12^{4} }\] thats your answer spring
anonymous
  • anonymous
yup
Jhannybean
  • Jhannybean
\[\frac{12^{-3}}{12^7} = 12^{-4} =\frac{1}{12^4} = \frac{1}{20736}\] TRy solving the next one yourself @Spring98
Spring98
  • Spring98
anonymous
  • anonymous
No problem

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