calculusxy
  • calculusxy
Help with exponents! Question posted below... MEDAL!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
calculusxy
  • calculusxy
\[\huge (x^4)^{-3} \times 2x^4\]
calculusxy
  • calculusxy
@hartnn
hartnn
  • hartnn
first term \(\huge (a^b)^c = a^{bc}\)

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More answers

hartnn
  • hartnn
you subtracted the exponents, the rule tell us to multiply them :3
anonymous
  • anonymous
@hartnn agree
calculusxy
  • calculusxy
Oh my godd
calculusxy
  • calculusxy
I am running our of my mind :\
calculusxy
  • calculusxy
*out
hartnn
  • hartnn
\(4 \times (-3) = - (4\times 3) = .. \)
calculusxy
  • calculusxy
\[\huge x^{4 \times (-3)} = x^{-12}\]
hartnn
  • hartnn
yussss
hartnn
  • hartnn
now \(\huge a^b \times a ^c = a^{b+c}\)
hartnn
  • hartnn
2 \((x^{-12} \times x^4) = .. ?\)
calculusxy
  • calculusxy
So here's the thing, I don't why we have to combine them... Is it because 2x^4 has two terms: 2 and x^4?
Jhannybean
  • Jhannybean
theres a common base, \(x\)
hartnn
  • hartnn
2 is a constant, lets throw that out of the process of multiplying 'x's
calculusxy
  • calculusxy
So just to clarify, if we have like 45x^5 then i will just have to take out the 45 and work with the x being the common base with another power right?
hartnn
  • hartnn
....with another term of x. yes.
calculusxy
  • calculusxy
\[\large 2(x^{-8}) = 2x^{-8}\]
calculusxy
  • calculusxy
Is that correct ?
hartnn
  • hartnn
\(2 x^3 \times 45 x^5 = 2\times 45 x^{3+5}\)
hartnn
  • hartnn
yes, thats correct :)
hartnn
  • hartnn
2/x^8 also "looks" nice
calculusxy
  • calculusxy
My answer is \[\huge \frac{ 2 }{ x^8 }\]
hartnn
  • hartnn
\(\huge \checkmark \)
calculusxy
  • calculusxy
The reason for why I have to put the x^8 to the denominator is because of the exponent of -8 right? The negative exponent tells the base to get to its reciprocal and then the exponent becomes positive?
hartnn
  • hartnn
correcto! \(\Large a^{-b} = \dfrac{1}{a^b}\)
calculusxy
  • calculusxy
Thank you!
hartnn
  • hartnn
welcome ^_^

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